Factorial Expression : Exercise

Factorials

We define $\boldsymbol{n} !=\boldsymbol{n}(\boldsymbol{n}-\mathbf{1})(\boldsymbol{n}-\mathbf{2}) \cdots \mathbf{3} \cdot \mathbf{2} \cdot \mathbf{1}$ if $n$ is a nonnegative integer. An empty product is normally defined to be 1 . With this convention, $0 !=1$ An alternative is to define $\boldsymbol{n} !$ recursively on the nonnegative integers. $\boldsymbol{n} != \begin{cases}1 & \text { if } \boldsymbol{n}=\mathbf{0} \\ \boldsymbol{n}(\boldsymbol{n}-\mathbf{1}) ! & \text { if } \boldsymbol{n} \geq \mathbf{1}\end{cases}$

Exercise

1. Evaluate
   $\begin{array}{lll} \text{(a)}\ 2 !& \text{(b)}\ 3 !& \text{(c)}\ 4 !\\\\ \text{(e)}\ 5 !& \text{(f)}\ 6 !& \text{(g)}\ 10 ! \end{array}$





$\begin{aligned} \text{(a)}\ &2 !=2 \times 1=2 \\\\ \text{(b)}\ &3 !=3 \times 2 \times 1=6 \\\\ \text{(c)}\ &4 !=4 \times 3 \times 2 \times 1=24 \\\\ \text{(d)}\ &5 !=5 \times 4 \times 3 \times 2 \times 1=120 \\\\ \text{(e)}\ &6 !=6 \times 5 \times 4 \times 3 \times 2 \times 1=720 \\\\ \text{(f)}\ &10 !=10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1=362880 \end{aligned}$



2. Express in factorial form:
   $\begin{array}{l} \text{(a)}\ 4 \times 3 \times 2 \times 1 \\\\ \text{(b)}\ 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \quad \\\\ \text{(c)}\ 6 \times 5 \\\\ \text{(d)}\ 8 \times 7 \times 6\\\\ \text{(e)}\ 10 \times 9 \times 8 \times 7 \\\\ \text{(f)}\ 15 \times 14 \times 13 \times 12\\\\ \text{(g)}\ \dfrac{9 \times 8 \times 7}{3 \times 2 \times 1} \\\\ \text{(h)}\ \dfrac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1}\\\\ \text{(i)}\ \dfrac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} \end{array}$





$\begin{aligned} \text{(a)}\ &\quad 4 \times 3 \times 2 \times 1\\\\ &=4 !\\\\ \text{(b)}\ &\quad 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\\\\ &=7 !\\\\ \text{(c)}\ &\quad 6 \times 5\\\\ &=\dfrac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1}\\\\ &=\dfrac{6 !}{4 !}\\\\ \text{(d)}\ &\quad 8 \times 7 \times 6\\\\ &=\dfrac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{5 \times 4 \times 3 \times 2 \times 1}\\\\ &=\dfrac{8 !}{5 !}\\\\ \text{(e)}\ &\quad 10 \times 9 \times 8 \times 7\\\\ &=\dfrac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1}\\\\ &=\dfrac{10 !}{6 !}\\\\ \text{(f)}\ &\quad 15 \times 14 \times 13 \times 12\\\\ &=\dfrac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}\\\\ &=\dfrac{15 !}{11 !}\\\\ \text{(g)}\ &\quad \dfrac{9 \times 8 \times 7}{3 \times 2 \times 1}\\\\ &=\dfrac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(6 \times 5 \times 4 \times 3 \times 2 \times 1)}\\\\ &=\dfrac{9 !}{3 ! 6 !}\\\\ \text{(h)}\ &\quad \dfrac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1}\\\\\ &=\dfrac{13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(4 \times 3 \times 2 \times 1)(9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}\\\\ &=\dfrac{13 !}{4 ! 9 !}\\\\ \text{(i)}\ &\quad \dfrac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1}\\\\ &=\dfrac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(5 \times 4 \times 3 \times 2 \times 1)( 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)}\\\\ &=\dfrac{15 !}{5 ! 10 !} \end{aligned}$ $\begin{aligned} &\textbf{Alternative Method }\\\\ \text{(a)}\ &\quad 4 \times 3 \times 2 \times 1\\\\ &=4 !\\\\ \text{(b)}\ &\quad 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\\\\ &=7 !\\\\ \text{(c)}\ &\quad 6 \times 5=\dfrac{6 \times 5 \times 4 !}{4 !}\\\\ &=\dfrac{6 !}{4 !}\\\\ \text{(d)}\ &\quad 8 \times 7 \times 6\\\\ &=\dfrac{8 \times 7 \times 6 \times 5 !}{5 !}\\\\ &=\dfrac{8 !}{5 !}\\\\ \text{(e)}\ &\quad 10 \times 9 \times 8 \times 7\\\\ &=\dfrac{10 \times 9 \times 8 \times 7 \times 6 !}{6 !}\\\\ &=\dfrac{10 !}{6 !}\\\\ \text{(f)}\ &\quad 15 \times 14 \times 13 \times 12=\dfrac{15 \times 14 \times 13 \times 12 \times 11 !}{11 !}\\\\ &=\dfrac{15 !}{11 !}\\\\ \text{(g)}\ &\quad \dfrac{9 \times 8 \times 7}{3 \times 2 \times 1}\\\\ &=\dfrac{9 \times 8 \times 7 \times 6 !}{3 \times 2 \times 1 \times 6 !}\\\\ &=\dfrac{9 !}{3 ! 6 !}\\\\ \text{(h)}\ &\quad \dfrac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1}\\\\ &=\dfrac{13 \times 12 \times 11 \times 10 \times 9 !}{4 \times 3 \times 2 \times 1 \times 9 !}\\\\ &=\dfrac{13 !}{4 ! 9 !} \\\\ \text{(i)}\ &\quad \dfrac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1}\\\\ &=\dfrac{15 \times 14 \times 13 \times 12 \times 11 \times 10 !}{5 \times 4 \times 3 \times 2 \times 1 \times 10 !}\\\\ &=\dfrac{15 !}{5 ! 10 !} \end{aligned}$



3. Simplify without using a calculator:
   $\begin{array}{ll} \text{(a)}\ \dfrac{7 !}{6 !}& \text{(b)}\ \dfrac{8 !}{6 !}\\\\ \text{(c)}\ \dfrac{12 !}{10 !}& \text{(d)}\ \dfrac{120 !}{119 !}\\\\ \text{(e)}\ \dfrac{10 !}{8 ! \times 2 !}& \text{(f)}\ \dfrac{100 !}{98 ! \times 2 !}\\\\ \text{(g)}\ \dfrac{7 !}{3 !}& \text{(h)}\ \dfrac{8 !}{5 !}\\\\ \text{(i)}\ \dfrac{4 !}{2 ! 2 !}& \text{(j)}\ \dfrac{6 !}{3 ! 2 !}\\\\ \text{(k)}\ \dfrac{6 !}{(3 !)^{2}}& \text{(l)}\ \dfrac{5 !}{3 !} \times \dfrac{7 !}{4 !} \end{array}$





$\begin{array}{l} \text{(a)}\ \dfrac{7 !}{6 !}=\dfrac{7 \times 6 !}{6 !}=7\\\\ \text{(b)}\ \dfrac{8 !}{6 !}=\dfrac{8 \times 7 \times 6 !}{6 !}=56\\\\ \text{(c)}\ \dfrac{12 !}{10 !}=\dfrac{12 \times 11 \times 10 !}{10 !}=132\\\\ \text{(d)}\ \dfrac{120 !}{119 !}=\dfrac{120 \times 119 !}{119 !}=120\\\\ \text{(e)}\ \dfrac{10 !}{8 ! \times 2 !}=\dfrac{10 \times 9 \times 8 !}{8 ! \times(2 \times 1)}=45\\\\ \text{(f)}\ \dfrac{100 !}{98 ! \times 2 !}=\dfrac{100 \times 99 \times 98 !}{98 ! \times 2 \times 1}=4950\\\\ \text{(g)}\ \dfrac{7 !}{3 !}=\dfrac{7 \times 6 \times 5 \times 4 \times 3 !}{3 !}=840\\\\ \text{(h)}\ \dfrac{8 !}{5 !}=\dfrac{8 \times 7 \times 6 \times 5 !}{5 !}=336\\\\ \text{(i)}\ \dfrac{4 !}{2 ! 2 !}=\dfrac{4 \times 3 \times 2 !}{2 !(2 \times 1)}=6\\\\ \text{(j)}\ \dfrac{6 !}{3 ! 2 !}=\dfrac{6 \times 5 \times 4 \times 3 !}{3 ! \times(2 \times 1)}=60\\\\ \text{(k)}\ \dfrac{6 !}{(3 !)^{2}}=\dfrac{6 \times 5 \times 4 \times 3 !}{3 !(3 \times 2 \times 1)}=20\\\\ \text{(l)}\ \dfrac{5 !}{3 !} \times \dfrac{7 !}{4 !}=(5 \times 4) \times(7 \times 6 \times 5)=4200 \end{array}$



4. Simplify:
   $\begin{array}{l} \text{(a)}\ \dfrac{n !}{(n-1) !}\\\\ \text{(b)}\ \dfrac{(n+2) !}{n !}\\\\ \text{(c)}\ \dfrac{(n+1) !}{(n-1) !} \end{array}$





$\begin{aligned} \text{(a)}\ &\dfrac{n !}{(n-1) !}\\\\\ &=\dfrac{n(n-1) !}{(n-1) !}\\\\ &=n \\\\ \text{(b)}\ &\dfrac{(n+2) !}{n !}\\\\ &=\dfrac{(n+2)(n+1) n !}{n !}\\\\ &=n^{2}+3 n+2 \\\\ \text{(c)}\ &\dfrac{(n+1) !}{(n-1) !}\\\\ &=\dfrac{(n+1) n(n-1) !}{(n-1) !}\\\\ &=n^{2}+n \end{aligned}$



5. Rewrite each of the following using factorial notation.
   $\begin{array}{l} \text{(a)}\ n(n-1)(n-2)(n-3)\\\\ \text{(b)}\ n(n-1)(n-2)(n-3)(n-4)(n-5)\\\\ \text{(c)}\ \dfrac{n(n-1)(n-2)}{5 \times 4 \times 3 \times 2 \times 1}\\\\ \text{(d)}\ \dfrac{n(n-1)(n-2)(n-3)(n-4)}{3 \times 2 \times 1} \end{array}$





$\begin{aligned} \text{(a)}\ &n(n-1)(n-2)(n-3)\\\\ &=\frac{n(n-1)(n-2)(n-3)(n-4) !}{(n-4) !}\\\\ &=\frac{n !}{(n-4) !} \\\\ \text{(b)}\ &n(n-1)(n-2)(n-3)(n-4)(n-5)\\\\ &=\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6) !}{(n-6) !}\\\\ &=\frac{n !}{(n-6) !} \\\\ \text{(c)}\ &\frac{n(n-1)(n-2)}{5 \times 4 \times 3 \times 2 \times 1}\\\\ &=\frac{n(n-1)(n-2)(n-3) !}{5 !(n-3) !}=\frac{n !}{5 !(n-3) !} \\\\ \text{(d)}\ &\frac{n(n-1)(n-2)(n-3)(n-4)}{3 \times 2 \times 1}\\\\ &=\frac{n(n-1)(n-2)(n-3)(n-4)(n-5) !}{3 !(n-5) !}\\\\ &=\frac{n !}{3 !(n-5) !} \end{aligned}$



6. Express the following as a single factorial notation.
   (a) $n !(n+1)$
   (b) $(n-1) !\left(n^{2}+n\right)$
   (c) $(n+4)(n+5)(n+3) !$
   (d) $n !\left(n^{2}+3 n+2\right)$
   (e) $(n+1)(n+2)(n+3)$
   (f) $(n-3)(n-4)(n-5)$





$\begin{aligned} \text { (a) } & n !(n+1) \\\\ &=(n+1) n ! \\\\ &=(n+1) ! \\\\ \text { (b) } &(n-1) !\left(n^{2}+n\right) \\\\ &=\left(n^{2}+n\right)(n-1) ! \\\\ &=(n+1) n(n-1) ! \\\\ &=(n+1) ! \\\\ &=(n+5)(n+4)(n+3) ! \\\\ &=(n+5) !\\\\ \text { (c) } &(n+4)(n+5)(n+3) ! \\\\ &=(n+5)(n+4)(n+3) ! \\\\ &=(n+5) !\\\\ \text { (d) } & n !\left(n^{2}+3 n+2\right) \\\\ =&(n+2)(n+1) n ! \\\\ =&(n+2) !\\\\ \text { (e) } &(n+1)(n+2)(n+3) \\\\ =& \frac{(n+3)(n+2)(n+1) n !}{n !} \\\\ =& \frac{(n+3) !}{n !}\\\\ \text { (f) } & (n-3)(n-4)(n-5) \\\\ &=\frac{(n-3)(n-4)(n-5)(n-6) !}{(n-6) !} \\\\ &=\frac{(n-3) !}{(n-6) !} \end{aligned}$



7. Write as a product by factorizing:
   (a) $5 !+4 !$
   (b) $11 !-10 !$
   (c) $5 !+7 !$
   (d) $12 !-10 !$
   (e) $9 !+8 !+7 !$
   (f) $7 !-6 !+8 !$
   (g) $12 !-2 \times 11 !$
   (h) $3 \times 9 !+5 \times 8 !$





$\begin{aligned} \text { (a) } & \quad 5 !+4 ! \\\\ &= 5 \times 4 !+4 ! \\\\ &=(5+1) 4 ! \\\\ &= 6 \times 4 ! \\\\ \text { (b) } & \quad 11 !-10 ! \\\\ &=(11-1) 10 ! \\\\ &= 10 \times 10 !\\\\ \text { (c) } & \quad 5 !+7 ! \\\\ &= 5 !+7 \times 6 \times 5 ! \\\\ &=(1+42) 5 ! \\\\ &= 43 \times 5 ! \\\\ \text { (d) } & \quad 12 !-10 ! \\\\ &= 12 \times 11 \times 10 !-10 ! \\\\ &=(132-1) 10 ! \\\\ &= 131 \times 10 !\\\\ \text { (e) } & \quad 9 !+8 !+7 ! \\\\ &= 9 \times 8 \times 7 !+8 \times 7 !+7 ! \\\\ &=(72+8+1) 7 ! \\\\ &= 81 \times 7 ! \\\\ \text { (f) } & \quad 7 !-6 !+8 ! \\\\ &= 7 \times 6 !-6 !+8 \times 7 \times 6 ! \\\\ &=(7-1+56) 6 ! \\\\ &= 62 \times 6 !\\\\ \text { (g) } & \quad 12 !-2 \times 11 ! \\\\ &= 12 \times 11 !-2 \times 11 ! \\\\ &=(12-2) 11 ! \\\\ &= 10 \times 11 ! \\\\ \text { (h) } & \quad 3 \times 9 !+5 \times 8 ! \\\\ &= 3 \times 9 \times 8 !+5 \times 8 ! \\\\ &=(27+5) 8 ! \\\\ &= 32 \times 8 ! \end{aligned}$



8. Simplify by factorizing:
   $\begin{array}{l} \text{(a)}\ \dfrac{12 !-11 !}{11}\\\\ \text{(b)}\ \dfrac{10 !+9 !}{11}\\\\ \text{(c)}\ \dfrac{10 !-8 !}{89}\\\\ \text{(d)}\ \dfrac{10 !-9 !}{9 !}\\\\ \text{(e)}\ \dfrac{6 !+5 !-4 !}{4 !}\\\\ \text{(f)}\ \dfrac{n !+(n-1) !}{(n-1) !}\\\\ \text{(g)}\ \dfrac{n !-(n-1) !}{n-1}\\\\ \text{(h)}\ \dfrac{(n+2) !+(n+1) !}{n+3} \end{array}$





$\begin{aligned} \text { (a) } & \quad \dfrac{12 !-11 !}{11} \\\\ &= \dfrac{12 \times 11 !-11 !}{11} \\\\ &= \dfrac{(12-1) 11 !}{11} \\\\ &= \dfrac{11 \times 11 !}{11} \\\\ &= 11 !\\\\ \text { (b) } & \quad \dfrac{10 !+9 !}{11} \\\\ &= \dfrac{10 \times 9 !+9 !}{11} \\\\ &= \dfrac{(10+1) 9 !}{11} \\\\ &= \dfrac{11 \times 9 !}{11} \\\\ &= 9 !\\\\ \text { (c) } & \quad \dfrac{10 !-8 !}{89} \\\\ &= \dfrac{10 \times 9 \times 8 !-8 !}{89} \\\\ &= \dfrac{(90-1) \times 8 !}{89} \\\\ &= \dfrac{89 \times 8 !}{89} \\\\ &= 8 !\\\\ \text { (d) } & \quad \dfrac{10 !-9 !}{9} \\\\ &= \dfrac{10 \times 9 !-9 !}{9} \\\\ &= \dfrac{(10-1) 9 !}{9} \\\\ &= \dfrac{9 \times 9 !}{9} \\\\ &= 9 !\\\\ \text { (e) } & \quad \dfrac{6 !+5 !-4 !}{4 !} \\\\ &= \dfrac{6 \times 5 \times 4 !+5 \times 4 !-4 !}{4 !} \\\\ &= \dfrac{(30+5-1) 4 !}{4 !} \\\\ &= 34\\\\ \text { (f) } & \quad \dfrac{n !+(n-1) !}{(n-1) !} \\\\ &=\dfrac{n(n-1) !+(n-1) !}{(n-1) !} \\\\ &=\dfrac{(n+1)(n-1) !}{(n-1) !} \\\\ &=n+1\\\\ \text { (g) } & \quad \dfrac{n !-(n-1) !}{n-1} \\\\ &=\dfrac{n(n-1) !-(n-1) !}{n-1} \\\\ &=\dfrac{(n-1)(n-1) !}{n-1} \\\\ &=(n-1) !\\\\ \text { (h) } & \quad \dfrac{(n+2) !+(n+1) !}{n+3} \\\\ &= \dfrac{(n+2)(n+1) !+(n+1) !}{n+3} \\\\ &= \dfrac{(n+2+1)(n+1) !}{n+3} \\\\ &= \dfrac{(n+3)(n+1) !}{n+3} \\\\ &=(n+1) ! \end{aligned}$

Indefinite Integration

Antiderivative ဟာ Grade 12 သင်ရိုးတွင် ပြဌာန်းလာမည့် သင်ခန်းစာတစ်ခုဖြစ်ပါတယ်။ သင်ခန်းစာရှင်းလင်းချက်များကို ဒီနေရာမှာ ရေခဲ့ဖူးပါတယ်။ အဆိုပါ post နှင့် ယှဉ်တွဲလေ့လာပြီး အောက်ပါ လေ့ကျင့်ခန်းများကို လေ့လာ လေ့ကျင့်ကြည့်နိုင်ပါတယ်။ စဉ်ဆက်မပြတ် လေ့လာသင်ယူနိုင်ကြပါစေ။

Rules of Integration

$\begin{array}{l} \text { 1. } \displaystyle\int k \ \mathrm{d}x=k x+c \\\\ \text { 2. } \displaystyle\int f^{\prime}(x) \ \mathrm{d}x=f(x)+c \\\\ \text { 3. } \displaystyle\int x^{n} \ \mathrm{d}x=\dfrac{x^{n+1}}{n+1}+c \\\\ \text { 4. } \displaystyle\int k f(x) \ \mathrm{d}x=k \displaystyle\int f(x) \ \mathrm{d}x\\\\ \text { 5. } \displaystyle\int\left[f(x) \pm g(x)\right] \ \mathrm{d}x = \displaystyle\int f(x) \ \mathrm{d}x \pm \displaystyle\int g(x) \ \mathrm{d}x \end{array}$

1. Integrate each of the following with respect to $x$.
   (a) $x^3$
   (b) $3\sqrt{x}$
   (c) $\dfrac{2}{x^2}$
   (d) $\dfrac{1}{2\sqrt{x}}$
   (e) $(3x+5)\ \mathrm{d}x$





$\begin{aligned} \text{(a)}\ \displaystyle\int x^3\ \mathrm{d}x &=\dfrac{x^{3+1}}{3+1}+C\\\\ &=\dfrac{x^4}{4}+C \end{aligned}$ $\begin{aligned} \text{(b)}\ \displaystyle\int 3\sqrt{x} \ \mathrm{d}x &=3\cdot \displaystyle\int \sqrt{x}\ \mathrm{d}x\\\\ &=3\cdot \displaystyle\int x^{\frac{1}{2}}\ \mathrm{d}x\\\\ &=3\cdot \dfrac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C\\\\ &=2x^{\frac{3}{2}}+C \end{aligned}$ $\begin{aligned} \text{(c)}\ \displaystyle\int \dfrac{2}{x^2}\ \mathrm{d}x &=2\cdot \displaystyle\int x^{-2}\ \mathrm{d}x\\\\ &=2\cdot \dfrac{x^{-2+1}}{-2+1} + C\\\\ &=-\dfrac{2}{x}+C \end{aligned}$ $\begin{aligned} \text{(d)}\ \displaystyle\int \dfrac{2}{x^2}\ \mathrm{d}x &=\dfrac{1}{2}\cdot \displaystyle\int \dfrac{1}{x^{\frac{1}{2}}}\ \mathrm{d}x\\\\ &=\dfrac{1}{2}\cdot \displaystyle\int x^{-\frac{1}{2}}\ \mathrm{d}x\\\\ &=\dfrac{1}{2}\cdot \dfrac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C\\\\ &=\sqrt{x}+C \end{aligned}$ $\begin{aligned} \text{(e)}\ \displaystyle\int (3x+5)\ \mathrm{d}x &= \displaystyle\int (3x)\ \mathrm{d}x+ \displaystyle\int (5)\ \mathrm{d}x\\\\ &= \frac{3x^2}{2}+5x+C \end{aligned}$



2. Find each of the following indefinite integrals.
   
   (a) $ \displaystyle\int (3x − 1)(x + 2)\ \mathrm{d}x$
   
   (b) $ \displaystyle\int\left(3 x^{3}-4 \sqrt{x}+3\right) \ \mathrm{d}x$
   
   (c) $ \displaystyle\int\left(6 x^{2}-\dfrac{4}{x^{2}}\right) \ \mathrm{d}x$
   
   (d) $ \displaystyle\int\left(5-\dfrac{1}{\sqrt{x}}+\dfrac{1}{x^{3}}\right) \ \mathrm{d}x$
   
   (e) $ \displaystyle\int \dfrac{x^{4}+5 x}{2 x^{3}} \ \mathrm{d}x$





$\begin{aligned} \text{(a)}\ \quad\quad &\displaystyle\int \left(3x-1\right)\left(x+2\right)\ \mathrm{d}x\\\\ =& \displaystyle\int (3x^2+5x-2)\ \mathrm{d}x\\\\ =& \displaystyle\int 3x^2\ \mathrm{d}x+ \displaystyle\int 5x\ \mathrm{d}x- \displaystyle\int 2\ \mathrm{d}x\\\\ =& x^3+\frac{5x^2}{2}-2x+C\\\\ \end{aligned}$ $\begin{aligned} \text{(b)}\quad\quad& \displaystyle\int\left(3 x^{3}-4 \sqrt{x}+3\right) \mathrm{d} x \\\\ =& \displaystyle\int 3 x^{3} \mathrm{~d} x-\displaystyle\int 4 \sqrt{x} \mathrm{~d} x+\displaystyle\int 3 \mathrm{~d} x \\\\ =& \displaystyle\int 3 x^{3} \ \mathrm{d}x-\displaystyle\int 4 x^{\frac{1}{2}} \ \mathrm{d}x+\displaystyle\int 3 \ \mathrm{d}x \\\\ =& \dfrac{3 x^{4}}{4}-\dfrac{4 x^{\frac{3}{2}}}{\frac{3}{2}}+3 x+C \\\\ =& \dfrac{3}{4} x^{4}-\dfrac{8}{3} x^{\frac{3}{2}}+3 x+C \end{aligned}$ $\begin{aligned} \text{(c)}\quad\quad & \displaystyle\int\left(6 x^{2}-\dfrac{4}{x^{2}}\right) \ \mathrm{d}x \\\\ =& \displaystyle\int 6 x^{2} \ \mathrm{d}x-\displaystyle\int \dfrac{4}{x^{2}} \ \mathrm{d}x \\\\ =& \displaystyle\int 6 x^{2} \ \mathrm{d}x-\displaystyle\int 4 x^{-2} \ \mathrm{d}x \\\\ =& \dfrac{6 x^{3}}{3}-\dfrac{4 x^{-1}}{-1}+C \\\\ =& 2 x^{3}+\dfrac{4}{x} \end{aligned}$ $\begin{aligned} \text{ (d) }\quad\quad & \displaystyle\int\left(5-\dfrac{1}{\sqrt{x}}+\dfrac{1}{x^{3}}\right) \mathrm{~d} x \\\\ =& \displaystyle\int 5 \mathrm{~d} x-\displaystyle\int x^{-\frac{1}{2}} \mathrm{~d} x+\displaystyle\int x^{-3} \mathrm{~d} x \\\\ =& 5 x-\dfrac{x^{\frac{1}{2}}}{\frac{1}{2}}+\frac{x^{-2}}{2}+C \\\\ =& 5 x-2 \sqrt{x}+\dfrac{1}{2 x^{2}}+C \end{aligned}$ $\begin{aligned} \text{(e)}\quad\quad& \displaystyle\int \dfrac{x^{4}+5 x}{2 x^{3}} \mathrm{~d} x \\\\ =& \displaystyle\int \dfrac{x^{4}}{2 x^{3}} \mathrm{~d} x+\displaystyle\int \dfrac{5 x}{2 x^{3}} \mathrm{~d} x \\\\ =& \displaystyle\int \dfrac{1}{2} x \mathrm{~d} x+\displaystyle\int \dfrac{5}{2} x^{-2} \mathrm{~d} x \\\\ =& \dfrac{1}{2} \cdot \dfrac{x^{2}}{2}+\dfrac{5}{2} \dfrac{x^{-1}}{-1}+C \\\\ =& \dfrac{1}{4} x^{2}-\dfrac{5}{2 x}+C \end{aligned}$



3. Find each of the following indefinite integrals.
   
   (a) $\displaystyle\int \dfrac{3 x}{2 \sqrt[5]{x^{2}}} \mathrm{~d} x$
   
   (b) $\displaystyle\int \dfrac{(3 x-1)^{2}}{5 x^{4}} \mathrm{~d} x$
   
   (c) $\displaystyle\int \dfrac{3 x^{7}+x^{2}}{2 \sqrt[3]{x}} \mathrm{~d} x$
   
   (d) $\displaystyle\int(x-3 \sqrt{x})^{2} \mathrm{~d} x$
   
   (e) $\displaystyle\int(1+\sqrt[4]{x})(1-\sqrt[4]{x}) \mathrm{d} x$
   
   (f) $\displaystyle\int\left(\sqrt[3]{x}+\dfrac{2}{\sqrt[3]{x}}\right)^{2} \mathrm{~d} x$





$\begin{aligned} \text { (a) } \quad\quad&\displaystyle\int \frac{3 x}{2 \sqrt[5]{x^{2}}} \mathrm{~d} x \\\\ =&\displaystyle\int \dfrac{3 x}{2 x^{\frac{2}{5}}} \mathrm{~d} x \\\\ =&\displaystyle\int \dfrac{3}{2} \cdot x^{-\frac{3}{2}} \mathrm{~d} x \\\\ =&\dfrac{3}{2} \frac{x^{-\frac{1}{2}}}{-\frac{1}{2}}+C \\\\ =&-\dfrac{3}{\sqrt{x}}+C \end{aligned}$ $\begin{aligned} \text { (b) } \quad\quad&\displaystyle\int \frac{(3 x-1)^{2}}{5 x^{4}} \mathrm{~d} x \\\\ =&\displaystyle\int \dfrac{9 x^{2}-6 x+1}{5 x^{4}} \mathrm{~d} x \\\\ =&\displaystyle\int \dfrac{9}{5} x^{-2} \mathrm{~d} x-\displaystyle\int \dfrac{6}{5} x^{-3} \mathrm{~d} x+\displaystyle\int \dfrac{1}{5} x^{-4} \mathrm{~d} x \\\\ =&\dfrac{9}{5} \dfrac{x^{-1}}{-1}-\dfrac{6}{5} \dfrac{x^{-2}}{-2}+\dfrac{1}{5} \dfrac{x^{-3}}{-3}+C \\\\ =&-\dfrac{9}{5 x}+\dfrac{6}{5 x^{2}}-\dfrac{1}{15 x^{3}}+C \end{aligned}$ $\begin{aligned} \text { (c) }\quad\quad& \displaystyle\int \dfrac{3 x^{7}+x^{2}}{2 \sqrt[3]{x}} \mathrm{~d} x \\\\ =&\displaystyle\int \dfrac{3 x^{7}}{2 \sqrt[3]{x}} \mathrm{~d} x+\displaystyle\int \dfrac{x^{2}}{2 \sqrt[3]{x}} \mathrm{~d} x \\\\ =&\displaystyle\int \dfrac{3}{2} \dfrac{x^{7}}{x^{\frac{1}{3}}} \mathrm{~d} x+\displaystyle\int \dfrac{1}{2} \dfrac{x^{2}}{x^{\frac{1}{3}}} \mathrm{~d} x \\\\ =&\displaystyle\int \dfrac{3}{2} x^{\frac{20}{3}} \mathrm{~d} x+\displaystyle\int \dfrac{1}{2} x^{\frac{5}{3}} \mathrm{~d} x\\\\ =&\dfrac{3}{2} \dfrac{x^{\frac{23}{3}}}{\frac{23}{3}}+\dfrac{1}{2} \dfrac{x^{\frac{8}{3}}}{\frac{8}{3}}+C \\ =&\dfrac{9}{46} x^{\frac{23}{3}}+\dfrac{3}{16} x^{\frac{8}{3}}+C \end{aligned}$ $\begin{aligned} \text { (d) } \quad\quad&\displaystyle\int(x-3 \sqrt{x})^{2} \mathrm{~d} x \\\\ =&\displaystyle\int\left(x^{2}-6 x \sqrt{x}+9 x\right) \mathrm{~d} x \\\\ =&\displaystyle\int x^{2} \mathrm{~d} x-\displaystyle\int 6 x^{\frac{3}{2}} \mathrm{~d} x+9 x \mathrm{~d} x \\\\ =&\dfrac{x^{2}}{3}-6 \dfrac{x^{\frac{3}{2}}}{\frac{3}{2}}+9 \dfrac{x^{2}}{2}+C \\\\ =&\dfrac{1}{3} x^{2}-4 x^{\frac{3}{2}}+\dfrac{9}{2} x^{2}+C \end{aligned}$ $\begin{aligned} \text { (e) } \quad\quad&\displaystyle\int(1+\sqrt[4]{x})(1-\sqrt[4]{x}) \mathrm{d} x \\\\ =&\displaystyle\int\left(1-(\sqrt[4]{x})^{2}\right) \mathrm{~d} x \\\\ =&\displaystyle\int\left(1-x^{\frac{1}{2}}\right) \mathrm{~d} x \\\\ =&\displaystyle\int 1 \mathrm{~d} x-\displaystyle\int x^{1 / 2} \mathrm{~d} x \\\\ =&x-\dfrac{x^{\frac{3}{2}}}{\frac{3}{2}}+C \\\\ =&x-\dfrac{2}{3} x^{\frac{3}{2}}+C \end{aligned}$ $\begin{aligned} \text { (f) } \quad\quad&\displaystyle\int\left(\sqrt[3]{x}+\frac{2}{\sqrt[3]{x}}\right)^{2} \mathrm{~d} x \\\\ =& \displaystyle\int\left(x^{\frac{1}{3}}+\dfrac{2}{x^{\frac{1}{3}}}\right)^{2} \mathrm{~d} x \\\\ =& \displaystyle\int\left(x^{\frac{2}{3}}+4+\dfrac{4}{x^{\frac{2}{3}}}\right) \mathrm{~d} x \\\\ =& \displaystyle\int x^{\frac{3}{2}} \mathrm{~d} x+\displaystyle\int 4 \mathrm{~d} x+\displaystyle\int 4 x^{-\frac{2}{3}} \mathrm{~d} x \\\\ =& \dfrac{x^{\frac{5}{2}}}{\frac{5}{2}}+4 x+4 \cdot \dfrac{x^{\frac{1}{3}}}{\frac{1}{3}}+C \\\\ =& \dfrac{2}{5} x^{\frac{5}{2}}+4 x+12 x^{\frac{1}{3}}+C \end{aligned}$



4. The rate of change of A with respect to r is given by $\dfrac{dA}{dr}= 4r+7$. If $A = 12$ when $r = 1$,find $A$ in terms of $r$.





$\begin{aligned} \dfrac{\mathrm{~d} A}{\mathrm{~d} r}=&4 r+7 \\\\ A=&\displaystyle\int \mathrm{~d} A \\\\ =&\displaystyle\int \frac{\mathrm{~d} A}{\mathrm{~d} r} \cdot \mathrm{~d} r \\\\ =&\displaystyle\int(4 r+7) \mathrm{~d} r\\\\ =&\displaystyle\int 4 r \cdot \mathrm{~d} r+\displaystyle\int 7 \mathrm{~d} r \\\\ =&2 r^{2}+7 r+C \\\\ \text { When }\ r=1,\ & A=12 \\\\ \therefore\ 12=&2+7+C \\\\ \therefore\ C=&3 . \\\\ \therefore\ A=&2 r^{2}+7 r+3 \end{aligned}$



5. Given that the gradient of a curve is $2x^2 + 7x$ and that the curve passes through the origin, determine the equation of the curve.





Let the given curve be $y$. $\begin{aligned} &\quad\text { Gradient of the curve } \\\\ &=\dfrac{\mathrm{~d}y}{\mathrm{~d}x}\\\\ &=2 x^{2}+7 x \\\\ &y=\displaystyle\int \mathrm{~d}y \\\\ &=\displaystyle\int \dfrac{\mathrm{~d}y}{\mathrm{~d}x} \mathrm{~d}x \\\\ &=\displaystyle\int\left(2 x^{2}+7 x\right) \mathrm{~d}x \\\\ &=\dfrac{2}{3} x^{3}+\dfrac{7}{2} x^{2}+C \end{aligned}$ through the origin $(0,0)$, $\begin{aligned} 0 &=\dfrac{3}{2}(0)^{3}+\dfrac{7}{2}(0)^{2}+C \\\\ \therefore C &=0 \end{aligned}$ Hence, the equation the curve is $y=\dfrac{2}{3} x^{3}+\dfrac{7}{2} x^{2}$



6. A curve is such that $\dfrac{dy}{dx}=k\sqrt[3]{x}$ , where $k$ is a constant and that it passes through the points $(1, 4)$ and $(8, 16)$. Find the equation of the curve.





$\begin{aligned} \dfrac{d y}{\mathrm{~d}x} &=k \sqrt[3]{x} \\\\ y &=\displaystyle\int \mathrm{~d}y \\\\ &=\displaystyle\int \dfrac{d y}{\mathrm{~d}x} \cdot \mathrm{~d}x \\\\ &=\displaystyle\int k \sqrt[3]{x} \mathrm{~d}x \\\\ &=k \displaystyle\int x^{\frac{1}{3}} \mathrm{~d}x \\\\ &=\dfrac{3 k}{4} x^{\frac{4}{3}}+C\\\\ \end{aligned}$ Since the curve passes through the point $(1,4)$ and $(8,16)$ $\begin{array}{l} 4=\dfrac{3 k}{4}+c\\\\ \therefore\ 3 k+4 C=16 \ldots(1)\\\\ 16=\dfrac{3 k}{4}(8)^{\frac{4}{3}}+C\\\\ 12 k+C=16\ldots(2) \end{array}$ Solving equatione $(1)$ and $(2)$, $\begin{array}{l} k=\dfrac{16}{15}, c=\dfrac{16}{5} \\\\ \therefore\ y=\dfrac{4}{5} x^{\frac{4}{3}}+\dfrac{16}{5} \end{array}$



7. The gradient of a curve at the point $(x, y)$ on the curve is given by $\dfrac{x^{2}-4}{x^{2}}$. Given that the curve passes through the point $(2,7)$, find the equation of the curve.





$\begin{aligned} \dfrac{d y}{\mathrm{~d}x} &=\dfrac{x^{2}-4}{x^{2}} \\\\ y &=\displaystyle \int \mathrm{~d}y \\\\ &=\displaystyle \int \dfrac{d y}{\mathrm{~d}x} \cdot \mathrm{~d}x \\\\ &=\displaystyle \int \dfrac{x^{2}-4}{x^{2}} \mathrm{~d}x \\\\ &=\displaystyle \int\left(1-\dfrac{4}{x^{2}}\right) \mathrm{~d}x \\\\ &=\displaystyle \int 1 \mathrm{~d}x-\displaystyle \int 4 x^{-2} \mathrm{~d}x \\\\ &=x+\dfrac{4}{x}+C \end{aligned}$ Since the curve passes through the point $(2,7)$, $7=2+\dfrac{4}{2}+C$ $C=3$ $\therefore\ y=x+\dfrac{4}{x}+3$



8. A curve with $\dfrac{dy}{dx}=k x+3$, where $k$ is a constant, passes through the point $P(3,19)$. Given that the gradient of the normal to the curve at the point $P$ is $-\dfrac{1}{15}$, find
   (i) the value of $k$,
   (ii) the equation of the curve,
   (iii) the coordinates of the turning point on the curve.
   





$\begin{aligned} \dfrac{\mathrm{~d}y}{\mathrm{~d}x}&=k x+3 \\\\ \text {gradient of normal at }&(3,19)=-\dfrac{1}{15} \\\\ \therefore-\left.\dfrac{1}{\dfrac{\mathrm{~d}y}{\mathrm{~d}x}}\right|_{(3,19)}&=-\dfrac{1}{15} \\\\ \left.\therefore \dfrac{1}{k x+3}\right|_{(3,19)}&=\dfrac{1}{15}\\\\ \therefore 3 k+3&=15 \\\\ k&=4 \\\\ \therefore \quad \dfrac{\mathrm{~d}y}{\mathrm{~d}x}&=4 x+3 \end{aligned}$ $\begin{aligned} y&=\displaystyle\int \mathrm{~d}y \\\\ &=\displaystyle\int \dfrac{\mathrm{~d}y}{\mathrm{~d}x} \cdot \mathrm{~d}x \\\\ &=\displaystyle\int(4 x+3) \mathrm{~d}x \\\\ &=2 x^{2}+3 x+C \end{aligned}$ Since the curve passe through the point $(3,19)$ $\begin{array}{l} 19 =2(3)^{2}+3(3)+C 19 =18+9+C \\\\ C =-8 \\\\ \therefore \quad\mathrm{~d}y=2 x^{2}+3 x-8\\\\ \text { At turning point, } \\\\ \dfrac{\mathrm{~d}y}{\mathrm{~d}x}=0 \\\\ 4 x+3=0\\\\ x=-\dfrac{3}{4} \\\\ y=2\left(-\dfrac{3}{4}\right)^{2}+3\left(-\dfrac{3}{4}\right)-8 \\\\ \quad =-\dfrac{73}{8} \\\\ \text { The turning point is } \left(-\dfrac{3}{4},-\dfrac{73}{8}\right) \end{array}$



9. The equation of a curve is such that $\dfrac{dy}{dx}=\dfrac{1}{(x-3)^{2}}+x .$ It is given that the curve passes through the point $(2,7)$. Find the equation of the curve.





$\begin{aligned} \dfrac{dy}{dx} &=\dfrac{1}{(x-3)^{2}}+x \\\\ y &=\displaystyle\int \mathrm{~d}y \\\\ &=\displaystyle\int \dfrac{dy}{dx} \\\\ &=\displaystyle\int\left(\dfrac{1}{(x-3)^{2}}+x\right) \mathrm{~d}x \\\\ &=\displaystyle\int \dfrac{1}{(x-3)^{2}} \mathrm{~d}x+\displaystyle\int x \mathrm{~d}x\\\\ \end{aligned}$ $\begin{aligned} \text{Since}\ \dfrac{d}{\mathrm{~d}x}(x-3)&=1, \\\\ d(x-3)&=\mathrm{~d}x \end{aligned}$ $\begin{aligned} y&=\displaystyle\int \dfrac{1}{(x-3)^{2}} d(x-3)+\displaystyle\int x \mathrm{~d}x \\\\ &=\displaystyle\int(x-3)^{-2} d(x-3)+\displaystyle\int x \mathrm{~d}x \\\\ &=-\dfrac{1}{x-3}+\dfrac{1}{2} x^{2}+c \end{aligned}$ Since $(2,7)$ lies on the curve, $\begin{array}{l} 7=-\dfrac{1}{2-3}+\dfrac{1}{2}(2)^{2}+C\\\\ C=4\\\\ \therefore\ y=\dfrac{1}{2} x^{2}-\dfrac{1}{x-3}+4 \end{array}$

Complex Number : Part (2)

Properties of Conjugate of Complex Number

Complex number $z=a+bi$ ၏ conjugate ဆိုသည်မှာ imaginary part ကို လက္ခဏာ ဆန့်ကျင်ဘက်သို့ ပြောင်းလိုက်ခြင်း ဖြစ်ပြီး သင်္ကေတအားဖြင့် $\overline{z}$ ဟုသတ်မှတ်ကြောင်း Part 1 တွင် တင်ပြခဲ့ပြီး ဖြစ်သည်။ ထိုကြောင့်

conjugate of $z$ $=a+bi = \overline{z}=a-bi$
ဟုဆိုနိုင်သည်။ $a$ သည် $z$ ၏ real part ဟုခေါ်ပြီး သင်္ကေတအားဖြင့် Re($z$) လည်းကောင်း၊ $b$ သည် $z$ ၏ imaginary part ဟုခေါ်ပြီး သင်္ကေတအားဖြင့် Im($z$) လည်းကောင်း၊ သတ်မှတ်ကြောင်း Part 1 တွင် တင်ပြခဲ့ပြီး ဖြစ်သည်။

အကယ်၍ $\text{Im}(z)=0$ ဖြစ်လျှင် $z=a + 0i = a =\text{Re}(z)$ ဖြစ်ပြီး $\overline{z}=a-0i = a=\text{Re}(z)$ ဖြစ်သွားမည်။ ထို့ကြောင့် မည့်သည့် complex number အတွက်မဆို အောက်ပါအဆိုသည် အမြဲမှန်ကန်သည်။

The relation $z = \overline{z}$ holds if and only if $z\in \mathbb{R}$.

Conjugate ၏ definition အရ $\overline{z}$ ၏ conjuagate $\overline{(\overline{z})}$ မှာ $\overline{z}$ ၏ imaginary part ကို လက္ခဏာပြောင်းပေးရန် ဖြစ်သည်။ ထို့ကြောင့်
conjugate of $\overline{z}$ $=\overline{(\overline{z})}=a - (-b)i= a+bi = z$
ဖြစ်သွားမည်။ ထို့ကြောင့် မည့်သည့် complex number အတွက်မဆို အောက်ပါအဆိုသည် အမြဲမှန်ကန်သည်။

For every complex number $z$, the relation $z = \overline{(\overline{z})}$ holds.

တဖန် $z=a+bi,\overline{z}=a-bi$ ဖြစ်သောကြောင့်

$\begin{aligned} z\cdot \overline{z} =&(a+bi)(a-bi)\\ z\cdot \overline{z} =& a^2-b^2i^2\\ =& a^2-b^2(-1)\\ =& a^2+b^2 \end{aligned}$

အထက်ပါ ရလဒ်အရ $z\cdot \overline{z}$ သည် အနုတ်မဟုတ်သော ကိန်းစစ်တစ်ခု ဖြစ်သည်ဟု ဆိုနိုင်သည်။ ထို့ကြောင့် မည့်သည့် complex number အတွက်မဆို အောက်ပါအဆိုသည် အမြဲမှန်ကန်သည်။

For every complex number $z$, the number $z\cdot \overline{z}\in \mathbb{R}$ is a nonnegative real number.

complex number နှစ်ခု $z_1=a+bi$ နှင့် $z_2=c+di$ ရှိသည် ဆိုပါစို့။ ထိုအခါ

$\begin{aligned} z_1+z_2 &=a+bi+c+di\\ &=(a+c)+(c+d)i\\ \overline{z_1+z_2} &= \overline{(a+c)+(c+d)i} \\ &= (a+c)-(c+d)i \\ \end{aligned}$

ဖြစ်မည်။တဖန်

$\begin{aligned} \overline{z_1}&=a-bi\\ \overline{z_2}&=c-di\\ \therefore\ \overline{z_1}+ \overline{z_2}&=a+c-bi-di\\ &= (a+c)-(c+d)i \\ \end{aligned}$

ရလဒ်အရ $\overline{z_1+z_2}=\overline{z_1}+ \overline{z_2}$ ဖြစ်သည်ဟုဆိုနိုင်သည်။ ထို့ကြောင့် မည့်သည့် complex number အတွက်မဆို အောက်ပါအဆိုသည် အမြဲမှန်ကန်သည်။

$\overline{z_1+z_2}=\overline{z_1}+ \overline{z_2}$ (The conjugate of a sum is the sum of the conjugates.)

ဆက်လက်၍ conjugate မြှောက်လဒ်များအကြောင်း လေ့လာပါမည်။

$\begin{aligned} z_1\cdot z_2 &=(a+bi)\cdot(c+di)\\ &=ac + adi + bci + bdi^2\\ &=ac + adi + bci - bd\\ &=(ac -bd) + (ad+ bc)i \\ \therefore\ \overline{z_1\cdot z_2}&= \overline{(ac -bd) + (ad+ bc)i} \\ \overline{z_1\cdot z_2}&= (ac -bd) - (ad+ bc)i \\ \end{aligned}$

$\begin{aligned} \overline{z_1}&=a-bi\\ \overline{z_2}&=c-di\\ \therefore\ \overline{z_1}\cdot \overline{z_2} &=(a-bi)\cdot(c-di)\\ &=ac - adi - bci + bdi^2\\ &=ac - adi - bci - bd\\ &=(ac - bd)-(ad + bc)i \\ \end{aligned}$

ဖော်ပြပါရလဒ်အရ $\overline{z_1\cdot z_2}=\overline{z_1}\cdot \overline{z_2}$ ဖြစ်သည်ဟုဆိုနိုင်သည်။ ထို့ကြောင့် မည့်သည့် complex number အတွက်မဆို အောက်ပါအဆိုသည် အမြဲမှန်ကန်သည်။

$\overline{z_1\cdot z_2}=\overline{z_1}\cdot \overline{z_2}$ (The conjugate of a product is the product of the conjugates.)

ဆက်လက်၍ conjugate စားလဒ်များအကြောင်း လေ့လာပါဦးမည်။

$\begin{aligned} \dfrac{z_1}{z_2}&=\dfrac{a+bi}{c+di}\\ &=\dfrac{a+bi}{c+di}\times \dfrac{c-di}{c-di}\\ &=\dfrac{(ac+bd) + (bc-ad)i}{c^2+d^2}\\ &=\dfrac{(ac+bd) + (bc-ad)i}{c^2+d^2}\\ &=\dfrac{ac +bd}{c^2+d^2} + \dfrac{bc-ad}{c^2+d^2}i\\ \therefore\ \overline{\left(\dfrac{z_1}{z_2}\right)}&=\dfrac{ac +bd}{c^2+d^2} - \dfrac{bc-ad}{c^2+d^2}i\\ \end{aligned}$

$\begin{aligned} \overline{z_1}&=a-bi\\ \overline{z_2}&=c-di\\ \therefore\ \dfrac{\overline{z_1}}{\overline{z_2}} &=\dfrac{a-bi}{c-di}\\ &=\dfrac{a-bi}{c-di}\times\dfrac{c+di}{c+di}\\ &=\dfrac{(ac+bd) - bci+adi}{c^2+d^2}\\ &=\dfrac{(ac+bd) - (bc-ad)i}{c^2+d^2}\\ &=\dfrac{ac +bd}{c^2+d^2} - \dfrac{bc-sad}{c^2+d^2}i\\ \end{aligned}$

ဖော်ပြပါရလဒ်အရ $\overline{\left(\dfrac{z_1}{z_2}\right)}=\dfrac{\overline{z_1}}{\overline{z_2}}$ ဖြစ်သည်ဟုဆိုနိုင်သည်။ ထို့ကြောင့် မည့်သည့် complex number အတွက်မဆို အောက်ပါအဆိုသည် အမြဲမှန်ကန်သည်။

$\overline{\left(\dfrac{z_1}{z_2}\right)}=\dfrac{\overline{z_1}}{\overline{z_2}}$ (The conjugate of a quotient is the quotient of the conjugates.)

မည့်သည့် complex number အတွက်မဆို အောက်ပါပုံသေနည်းသည် အမြဲမှန်ကန်သည်။ သက်သေပြချက်ကို စာဖတ်သူကိုယ်တိုင် သက်သေပြကြည့်ပါ။

The formulas
$\operatorname{Re}(z)=\dfrac{z+\bar{z}}{2} \text { and } \operatorname{Im}(z)=\dfrac{z-\bar{z}}{2 i} $
are valid for all $z \in \mathbb{C}.$

Power of $\mathbf{i}$

By definition,

$\begin{array}{l} \hline i^1 =i \\ i^2 =-1\\ i^3 =i^2\cdot i = -1\cdot i=-i\\ i^4 =i^3\cdot i = -i\cdot i=-i^2=-(-1)=1\\ \hline i^5 =i^4\cdot i = 1\cdot i=i\\ i^6 =i^5\cdot i = i\cdot i=i^2=-1\\ i^7 =i^6\cdot i = -1\cdot i=-i\\ i^8 =i^7\cdot i = -i\cdot i=-i^2=-(-1)=1\\ \hline \end{array}$ The cycle is repeated continuously: $i$, $−1$, $− i$, $1$, every four powers.

အထက်ပါဖွဲ့စည်းပုံကို လေ့လာခြင်းအားဖြင့် $i$ ၏ ထပ်ညွှန်းသည် $4$ ၏ ဆတိုးကိန်းဖြစ်တိုင်း $1$ နှင့် ညီကြောင်းတွေ့ရသည်။ ထို့ကြောင့် $i^4$ = $i^8$ = $i^{12}$ = ... = $i^{4n}$ = $1$ ဟု ပုံသေ မှတ်ယူနိုင်ပါသည်။

Example 1 Simplify each power of $i$. (a) $i^{13}\quad$ (b) $i^{56}\quad$ (c) $i^{-3}$ Solution $\begin{array}{ll} \text{(a)}\quad i^{13} &=i^{12+1}\\ &=i^{12}\cdot i\\ &= (i^{4})^3\cdot i\\ &= (1)^3\cdot i\\ &= i\\ \text{(b)}\quad i^{56} &= (i^{4})^{14}\\ &= 1^{14}\\ &= 1\\ \text{(c)}\quad i^{-3} &= i^{-4+1}\\ &= i^{-4}\cdot i\\ &= (i^{4})^{-1}\cdot i\\ &= (1)^{-1}\cdot i\\ &= 1\cdot i\\ &= i \end{array}$

Example 2 Simplify the complex number and write it in standard form. (a) $-3i^{5} + 2i^2\quad$ (b) $(\sqrt{-32})^3\quad$ (c) $\dfrac{1}{5i^{3}}$ Solution $\begin{array}{ll} \text{(a)}\quad -3 i^{5}+2 i^{2} &=-3 i^{4} \cdot i+2(-1) \\ &=-3(1) \cdot i-2 \\ &=-2-3 i \\ \text{(b)}\quad (\sqrt{-32})^3 &=(4 \sqrt{2} \sqrt{-1})^{3} \\ &=(4 \sqrt{2} i)^{3} \\ &=64(2 \sqrt{2}) i^{3}\\ &=128 \sqrt{2} i^{2} \cdot i \\ &=128 \sqrt{2}(-1) i \\ &=-128 \sqrt{2} i \\ \text{(c)}\quad \dfrac{1}{5 i^{3}} &=\dfrac{1}{5} i-3 \\ &=\dfrac{1}{5} i^{-4} \cdot i \\ &=\dfrac{1}{5}\left(i^{4}\right)^{-1} i \\ &=\dfrac{1}{5}(1)^{-1} i \\ &=\dfrac{1}{5} i \end{array}$

Complex Solutions of a Quadratic Equation

Quadratic Equation တစ်ခု၏ standard equation form မှာ

• $ax^2+bx+c=0$ ဖြစ်ကြောင်းသိရှိခဲ့ပြီး ဖြစ်ကြောင်း
• Quadratic Equation ကိုပြေလည်စေသော solution ကို quadratic formula $ \dfrac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}$ ဖြင့်ရှာယူနိုင်ကြောင်း
• $b^2-4ac$ ကို quadratic function တစ်ခု၏ discriminant ဟု ခေါ်ကြောင်း
• discriminant < 0 ဖြစ်ပါက ပေးထားသော quadratic equation ကို ပြေလည်စေသော ကိန်းစစ် အဖြေမရှိကြောင်း

သိရှိခဲ့ပြီး ဖြစ်သည်။

သို့သော် discriminant < 0 ဖြစ်သော quadratic equation များအတွက် complex solution ကို ရှာယူနိုင်ပါသည်။ အောက်ပါ ဥပမာများကို လေ့လာကြည့်ပါ။

Example 3 Verify that the equation $x^2+5x+7=0$ has no real solution, hence find the complex solutions. Solution $x^2+5x+7=0$ Comparing with $ax^2+bx+c=0$, we have $a=1$, $b=5$ and $c=7$. $\therefore\quad b^2-4ac=5^2-4(1)(7)=-3 < 0$ Since the discriminant < 0, there is no real solution for the given equation. $\begin{aligned} x &= \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\ &= \dfrac{-5\pm \sqrt{-3}}{2}\\ &= \dfrac{-5}{2}\pm \dfrac{\sqrt{-3}}{2}\\ &= \dfrac{-5}{2}\pm \dfrac{\sqrt{3}\sqrt{-1}}{2}\\ &= \dfrac{-5}{2}\pm \dfrac{\sqrt{3}}{2}i \end{aligned}$

Argand diagram (Complex Plane)

Complex Number များကို complex plane တွင် နေရချသတ်မှတ်နိုင်ကြောင်း Part (1) တွင် တင်ပြခဲ့ပြီး ဖြစ်သည်။ အဆိုပါ complex plane ကို Argand plane (သို့မဟုတ်) Gauss Plane ဟုလည်း ခေါ်သည်။ $z=a+bi$ ၏ တည်နေရာသည် Argand plane တွင် $P(a,b)$ ဖြစ်သည်။ ထို့ကြောင့်

• $z_1=4+3i$ တည်နေရာသည် Argand plane တွင် $(4,3)$
• $z_2=3-2i$ တည်နေရာသည် Argand plane တွင် $(3,-2)$
• $z_3=-1-5i$ တည်နေရာသည် Argand plane တွင် $(-1,-5)$
• $z_4=-4+3i$ တည်နေရာသည် Argand plane တွင် $(-4,3)$
• $z_5=3i$ တည်နေရာသည် Argand plane တွင် $(0,3)$
• $z_6=-2i$ တည်နေရာသည် Argand plane တွင် $(0,-2)$
• $z_7=-5$ တည်နေရာသည် Argand plane တွင် $(-5,0)$
• $z_8=3$ တည်နေရာသည် Argand plane တွင် $(3,0)$

အသီးသီး ဖြစ်ကြသည်။

ထို့ကြောင့် complex number တစ်ခုကို argand diagram (complex plane) ပေါ်ရှိ coordinate ဖြင့်လည်း ဖေါ်ပြလေ့ရှိသည်။ ဥပမာ

$\begin{array}{lll} z_1=4+3i & \text{or} & z_1=(4,3)\\ z_2=3-2i & \text{or} & z_2=(3,-2)\\ z_3=-1-5i & \text{or} & z_3=(-1,-5)\\ z_4=-4+3i & \text{or} & z_4=(-4,3)\\ z_5=3i & \text{or} & z_5=(0,3)\\ z_6=-2i & \text{or} & z_6=(0,-2)\\ z_7=-5 & \text{or} & z_7=(-5,0)\\ z_8=3 & \text{or} & z_8=(3,0)\\ \end{array}$

ဟုဖေါ်ပြနိုင်သည်။

Modulus and Argument of a Complex Number

Complex number $z =x+yi$ ကို Complex plane တွင် plot လုပ်သည့်အခါ အောက်ပါအတိုင်း ရသည် ဆိုပါစို့။

ထိုအခါ origin မှ ပေးထားသော အမှတ်ကိုဆက်သော မျဉ်းပြတ်၏ အလျားကို $|z|$ ဟုခေါ်သည်။ Pythagoras theorem အရ $|z|=\sqrt{x^2+y^2}$ ဟု အလွယ်တကူသိနိုင်သည်။ အဆိုပါမျဉ်းပြတ်နှင့် positive x-axis ကြားရှိထောင့် ($\theta$) ကို argument of $z$ ဟုခေါ်ပြီး သင်္ကေတအားဖြင့် $\operatorname{arg}(z)$ဟုခေါ်သည်။

$\begin{array}{l} \text{By Pythagoras' Theorem}\\ |z|=\sqrt{x^2+y^2}\\ \text{Since}\ \tan{\theta}=\dfrac{y}{x},\\ \theta=\tan^{-1}\dfrac{y}{x}\\ \operatorname{arg}(z)=\tan^{-1}\left(\dfrac{y}{x}\right) \end{array}$

$\operatorname{arg}(z)=\theta$ ကို ဖော်ပြရာတွင် $-180^{\circ} < \theta \le 180^{\circ}$ (radian ဖြင့်ဖော်ပြသော် $-\pi < \theta \le \pi$) interval ဖြင့် ဖော်ပြသည်။ ၎င်းကို principal argument ဟုလည်းခေါ်သည်။ ဆိုလိုသည်မှာ ပေးထားသော complex number သည် argand plane ၏ first နှင့် second quadrant တွင်ရှိပါက $0<\operatorname{arg}(z)<180^{\circ}$ (positive angle) ဖြစ်ပြီး ပေးထားသော complex number သည် argand plane ၏ third နှင့် fourth quadrant တွင်ရှိပါက $-180^{\circ}<\operatorname{arg}(z)< 0$ (negative angle) ဖြစ်မည်။

$\begin{array}{cccc} \hline z & |z| & \text{location} & \text{argument}\\ \hline a+bi\\(a,b>0) &\sqrt{a^2+b^2} & 1^{\text{st}}\text{quadrant} & \operatorname{arg}(z)=\tan^{-1}\left|\dfrac{b}{a}\right|\\ \hline a+bi\\(a<0,b>0) &\sqrt{a^2+b^2} & 2^{\text{nd}} \text{quadrant} & \operatorname{arg}(z)=\left(180^{\circ}-\tan^{-1}\left|\dfrac{b}{a}\right|\right)\\ \hline a+bi \\(a,b<0) &\sqrt{a^2+b^2} & 3^{\text{rd}} \text{quadrant} & \operatorname{arg}(z)=-\left(180^{\circ}-\tan^{-1}\left|\dfrac{b}{a}\right|\right)\\ \hline a+bi\\(a>0,b<0) & \sqrt{a^2+b^2}& 4^{\text{th}} \text{quadrant} &\operatorname{arg}(z)=-\tan^{-1}\left|\dfrac{b}{a}\right|\\ \hline a \\(a>0) & |a| & \text{on positive}\ x\text{-axis} & 0^{\circ}\\ \hline a \\(a<0) & |a| & \text{on negative}\ x\text{-axis} & 180^{\circ}\\ \hline bi\\(b>0) & |b| & \text{on positive}\ y\text{-axis} & 90^{\circ}\\ \hline bi \\(b<0)& |b| &\text{on negative}\ y\text{-axis} & -90^{\circ}\\ \hline \end{array}$

Example 4 Plot the following complex number in complex plane and hence find the modulus and argument of each number. (a) $u=3+4i$ (b) $v=-1+i$ (c) $w=-2-3i$ (d) $z=3-5i$ Solution (a) $\quad u=3+4i$ $\quad\quad |u|=\sqrt{3^2+4^2}=5$ $\quad\quad \tan^{-1}\left(\dfrac{4}{3}\right)=53.13^{\circ}$ $\therefore\quad \operatorname{arg}(u)=53.13^{\circ}$ (b) $\quad v=-1+i$ $\quad\quad |v|=\sqrt{(-1)^2+1^2}=\sqrt{2}$ $\quad\quad \tan^{-1}\left(\dfrac{1}{1}\right)=45^{\circ}$ $\therefore\quad \operatorname{arg}(v)=(180^{\circ}-45^{\circ})=135^{\circ}$ (c) $\quad w=-2-3i$ $\quad\quad |w|=\sqrt{(-2)^2+(-3)^2}=\sqrt{13}$ $\quad\quad \tan^{-1}\left(\dfrac{3}{2}\right)=56.31^{\circ}$ $\therefore\quad \operatorname{arg}(w)=-(180^{\circ}-56.31^{\circ})=-123.69^{\circ}$ (d) $\quad z=3-5i$ $\quad\quad |z|=\sqrt{(3)^2+(-5)^2}=\sqrt{34}$ $\quad\quad \tan^{-1}\left(\dfrac{5}{3}\right)=59.04^{\circ}$ $\therefore\quad \operatorname{arg}(z)=-59.04^{\circ}$

$z=x+yi$ ဖြစ်သည်ဆိုပါစို့။ ထိုအခါ $\overline{z}=x-yi$ ဖြစ်မည်။ ထို့ကြောင့်

$z\cdot\overline{z} = x^2+y^2 = |z|^2$

ဖြစ်မည်။ ထို့ကြောင့် မည်သည့် complex number အတွက်မဆို အောက်ပါအဆို သည် အမြဲမှန်ကန်သည်။

For every $z\in \mathbb{C}, z\cdot\overline{z} =|z|^2$.

Example 5 If $\operatorname{Im}\left(\dfrac{z+i}{z-i}\right)=0$, where $z \neq i$, show that $\operatorname{Re}(z)=0$. Solution $\begin{aligned} \text { Let } w &=\dfrac{z+i}{z-i} \\\\ \overline{w} &= \overline{\left(\dfrac{z+i}{z-i}\right)} \\\\ &=\dfrac{\overline{z+i}}{\overline{z-i}} \\\\ &=\dfrac{\overline{z}+\overline{i}}{\overline{z}-\overline{i}} \\\\ &=\dfrac{\overline{z}-i}{\overline{z}+i}\\\\ \therefore\ w-\overline{w} &=\dfrac{z+i}{z-i}-\dfrac{\overline{z}-i}{\overline{z}+i} \\\\ &=\dfrac{(z+i)(\overline{z}+i)-(z-i)(\overline{z}-i)}{(z-i)(\overline{z}+i)} \\\\ &=\dfrac{z \overline{z}+z i+\overline{z} i-1-z \overline{z}+z i+\overline{z}^{i}+1}{(z-i)(\overline{z}+i)} \\\\ &=\dfrac{2(z+\overline{z}) i}{(z-i)(\overline{z}+i)}\\\\ \dfrac{w-\overline{w}}{2 i}&=\dfrac{z+\overline{z}}{(z-i)(\overline{z}+i)} \\\\ \operatorname{Im}(w)&=\dfrac{z+\overline{z}}{(z-i)(\overline{z}+i)} \\\\ \text { By the}\ & \text{problem, } \\\\ \text { Im }(w)&=0 \\\\ \dfrac{z+\overline{z}}{(z-i)(\overline{z}+i)}&=0\\\\ \therefore \quad z+\overline{z}&=0 \\\\ \quad\quad \dfrac{z+\overline{z}}{2}&=0 \\\\ \therefore\quad \operatorname{Re}(z)&=0 \end{aligned}$

Example 6 Given that $z \in \mathbb{C}$ and $|z|=1$, show that $\operatorname{Re}\left(\dfrac{z-1}{z+1}\right)=0$. Solution $\begin{aligned} \text { Let } &\frac{z-1}{z+1}=w \\\\ \therefore \overline{w} &=\overline{\left(\frac{z-1}{z+1}\right)} \\\\ &=\frac{\overline{z-1}}{\overline{z+1}} \\\\ &=\frac{\overline{z}-1}{\overline{z}+1} \\\\ w+\overline{w}&=\frac{z-1}{z+1}+\frac{\overline{z}-1}{\overline{z}+1}\\\\ w+\overline{w} &=\dfrac{z \overline{z}+z-\overline{z}-1+z \overline{z}-z+\overline{z}-1}{(z+1)(\overline{z}+1)} \\\\ w+\overline{w} &=\dfrac{2(z \overline{z}-1)}{(z+1)(\overline{z}+1)} \\\\ \therefore \quad \dfrac{w+\overline{w}}{2} &=\dfrac{z \overline{z}-1}{(z+1)(\overline{z}+1)} \\\\ \operatorname{Re}(w) &=\dfrac{|z|^{2}-1}{(z+1)(\overline{z}+1)}\\\\ &=0 \quad(\because|z|=1) \\\\ \therefore \operatorname{Re}\left(\dfrac{z-1}{z+1}\right)&=0 . \end{aligned}$

Example 7 Given that $z^{2}=15+8i$ and and $z=x+y$ i where $z \in \mathbb{C}$ and $x$ and $y$ are integers to be found. Hence find $\sqrt{15+8 i}.$ Using information frorm $(1)$, solve the equation $w^{2}-(2+3 i) w-5+i=0$ for $w\in \mathbb{C}$. Solution $\begin{aligned} 1.\quad\quad\quad\quad\quad\quad z & =x+y i \\\\ z^{2} & =15+8 i \\\\ (x+y i)^{2} & =15+8 i \\\\ x^{2}+2 x y i+y^{2} i^{2}&=15+8 i \\\\ \therefore \quad x^{2}-y^{2}& =15\quad \ldots(1) \\\\ \quad\quad 2 x y &=8 \quad \ldots(2)\\\\ \therefore x y=4 & \Rightarrow y=\dfrac{4}{x} \\\\ \therefore x^{2}-\left(\dfrac{16}{x^{2}}\right)&=15 \\\\ x^{4}-16&=15 x^{2} \\\\ x^{4}-15 x^{2}-16&=0 \\\\ \left(x^{2}-16\right)\left(x^{2}+1\right)&=0 \\\\ x^{2}=16\ \text { or }\ & x^{2}=-1 \\\\ \therefore x=\pm 4\ \text { or }\ & x=i \end{aligned}$ $\text { Since }\ x \text { is an integer, } x=\pm 4$ When $x=4, y=\dfrac{4}{4}=1$ When $x=-4, y=\dfrac{4}{-4}=-1$ $\quad z=4+i \text { (or) } z=-(4+i)$ $\begin{aligned} \therefore \quad \sqrt{15+8 i} &=\sqrt{z^{2}} \\\\ &=z \\\\ &=\pm(4+i) \end{aligned}$ $\begin{array}{l} \text { 2. } w^{2}-(2+3 i) w-5+i=0 \\\\ \text { Comparing with } a w^{2}+b w+c =0, \\\\ \quad a=1, b=-(2+3 i) \text { and } c=-5+i \\\\ \therefore\quad w =\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\\\ \quad\quad\quad =\dfrac{2+3 i \pm \sqrt{(2+3 i)^{2}-4(-5+i)}}{2} \\\\ \quad\quad\quad =\dfrac{2+3 i \pm \sqrt{4+12 i-9+20-4 i}}{2}\\\\ \quad\quad\quad =\dfrac{2+3 i \pm \sqrt{15+8 i}}{2} \\\\ \quad\quad\quad =\dfrac{2+3 i \pm(4+i)}{2} \\\\ \therefore \quad w=\dfrac{2+3 i+4+i}{2}=3+2 i \\\\ \quad\quad\quad\quad(\text { or }) \\\\ \quad\quad w=\dfrac{2+3 i-4-i}{2}=-1+i \end{array}$

Example 8 The opposition to current in an electrical circuit is called its impedance. The impedance $z$ in a parallel circuit with two pathways satisfies the equation $\dfrac{1}{z}=\dfrac{1}{z_{1}}+\dfrac{1}{z_{2}}$ where $z_1$ is the impedance (in ohms) of pathway 1 and $z_2$ is the impedance (in ohms) of pathway 2. The impedance of each pathway in a parallel circuit is found by adding the impedances of all components in the pathway. Use the table to find $z_1$ and $z_2$. Find the impedance $z$. Solution $\begin{aligned} z_{1}&=9+16 i \text { ohm } \\\\ z_{2}&=20-10 i \text { ohm } \\\\ \therefore\quad z_{1}+z_{2}&=29+6 i \text { ohm } \\\\ z_{1} \cdot z_{2}& =340+230 i \text { ohm } \\\\ \dfrac{1}{z}&=\dfrac{1}{z_{1}}+\dfrac{1}{z_{2}} \\\\ &=\dfrac{z_{1}+z_{2}}{z_{1} z_{2}}\\\\ z &=\dfrac{z_{1} z_{2}}{z_{1}+z_{2}} \\\\ &=\dfrac{340+230 i}{29+6 i} \times \dfrac{29-6 i}{29-6 i} \\\\ &=\dfrac{11240+4630 i}{877} \\\\ &=\dfrac{11240}{877}+\dfrac{4630}{877} i \text { ohm } \end{aligned}$

Exercises Find the modulus and argument of each of the following complex numbers in the complex plane below. Plot the following complex numbers in single complex plane and find the the modulus and argument of each complex number $z_{1}=3+i $ $z_{2}=-4+2 i $ $z_{3}=-5-4 i $ $z_{4}=5-i$ $z_{5}=1 $ $z_{6}=-3 i$ $z_{7}=2 i $ $z_{8}=-4$ Solve the equations: $x^{2}+9=0$ $9 x^{2}+25=0$ $x^{2}+2 x+2=0$ $x^{2}+x+1=0$ $2 x^{2}+3 x+2=0$ Find the quadratic equation which has roots $2 \pm \sqrt{3} i$ Find the value of the real number $y$ such that $(3+2 i)(1+i y)$ is (a) real (b) imaginary. Simplify $\dfrac{1+i}{1-i}-(1+2 i)(2+2 i)+\dfrac{3-i}{1+i}$ $2 i(i-1)+(\overline{\sqrt{3}+i})^{3}+(1+i) \overline{(1+i)}$ Find $a \in \mathbb{R}$ such that $z=-i$ is a root for the polynomial $P(z)=z^{3}-z^{2}+z+1+a .$ Hence find the factors of $P(z)$ in $\mathbb{R}$ and in $\mathbb{C}$. Solve each of the following equations for the complex number $z$. $4+5 i=z-(1-i)$ $(1+2 i) z=2+5 i$ $z(2+i)=3-2 i$ $(z+i)(1-i)=2+3 i$ $\dfrac{1}{z}+\dfrac{1}{2-i}=\dfrac{3}{1+i}$