Complex number $z=a+bi$ ၏ conjugate αိုαα်αှာ imaginary part αို αα္ααာ αα့်αျα်αα်αို့ αြောα်းαိုα်αြα်း αြα ်αြီး αα်္αေαα‘ားαြα့် $\overline{z}$ αုαα်αှα်αြောα်း Part 1 αွα် αα်αြαဲ့αြီး αြα ်αα်။ αိုαြောα့်
αုαိုαိုα်αα်။ $a$ αα် $z$ ၏ real part αုαေါ်αြီး αα်္αေαα‘ားαြα့် Re($z$) αα်းαောα်း၊ $b$ αα် $z$ ၏ imaginary part αုαေါ်αြီး αα်္αေαα‘ားαြα့် Im($z$) αα်းαောα်း၊ αα်αှα်αြောα်း Part 1 αွα် αα်αြαဲ့αြီး αြα ်αα်။
α‘αα်၍ $\text{Im}(z)=0$ αြα ်αျှα် $z=a + 0i = a =\text{Re}(z)$ αြα ်αြီး $\overline{z}=a-0i = a=\text{Re}(z)$ αြα ်αွားαα်။ αို့αြောα့် αα့်αα့် complex number α‘αွα်ααို α‘ောα်αါα‘αိုαα် α‘αြဲαှα်αα်αα်။
The relation $z = \overline{z}$ holds if and only if $z\in \mathbb{R}$.
Conjugate ၏ definition α‘α $\overline{z}$ ၏ conjuagate $\overline{(\overline{z})}$ αှာ $\overline{z}$ ၏ imaginary part αို αα္ααာαြောα်းαေးαα် αြα ်αα်။ αို့αြောα့်
αြα ်αွားαα်။ αို့αြောα့် αα့်αα့် complex number α‘αွα်ααို α‘ောα်αါα‘αိုαα် α‘αြဲαှα်αα်αα်။
For every complex number $z$, the relation $z = \overline{(\overline{z})}$ holds.
ααα် $z=a+bi,\overline{z}=a-bi$ αြα ်αောαြောα့်
α‘αα်αါ ααα်α‘α $z\cdot \overline{z}$ αα် α‘αုα်ααုα်αော αိα်းα α ်αα ်αု αြα ်αα်αု αိုαိုα်αα်။ αို့αြောα့် αα့်αα့် complex number α‘αွα်ααို α‘ောα်αါα‘αိုαα် α‘αြဲαှα်αα်αα်။
For every complex number $z$, the number $z\cdot \overline{z}\in \mathbb{R}$ is a nonnegative real number.complex number αှα ်αု $z_1=a+bi$ αှα့် $z_2=c+di$ αှိαα် αိုαါα ို့။ αိုα‘αါ
αြα ်αα်။ααα်
ααα်α‘α $\overline{z_1+z_2}=\overline{z_1}+ \overline{z_2}$ αြα ်αα်αုαိုαိုα်αα်။ αို့αြောα့် αα့်αα့် complex number α‘αွα်ααို α‘ောα်αါα‘αိုαα် α‘αြဲαှα်αα်αα်။
$\overline{z_1+z_2}=\overline{z_1}+ \overline{z_2}$ (The conjugate of a sum is the sum of the conjugates.)
αα်αα်၍ conjugate αြှောα်αα်αျားα‘αြောα်း αေ့αာαါαα်။
αော်αြαါααα်α‘α $\overline{z_1\cdot z_2}=\overline{z_1}\cdot \overline{z_2}$ αြα ်αα်αုαိုαိုα်αα်။ αို့αြောα့် αα့်αα့် complex number α‘αွα်ααို α‘ောα်αါα‘αိုαα် α‘αြဲαှα်αα်αα်။
$\overline{z_1\cdot z_2}=\overline{z_1}\cdot \overline{z_2}$ (The conjugate of a product is the product of the conjugates.)
αα်αα်၍ conjugate α ားαα်αျားα‘αြောα်း αေ့αာαါα¦းαα်။
αော်αြαါααα်α‘α $\overline{\left(\dfrac{z_1}{z_2}\right)}=\dfrac{\overline{z_1}}{\overline{z_2}}$ αြα ်αα်αုαိုαိုα်αα်။ αို့αြောα့် αα့်αα့် complex number α‘αွα်ααို α‘ောα်αါα‘αိုαα် α‘αြဲαှα်αα်αα်။
$\overline{\left(\dfrac{z_1}{z_2}\right)}=\dfrac{\overline{z_1}}{\overline{z_2}}$ (The conjugate of a quotient is the quotient of the conjugates.)
αα့်αα့် complex number α‘αွα်ααို α‘ောα်αါαုံαေαα်းαα် α‘αြဲαှα်αα်αα်။ αα်αေαြαျα်αို α ာαα်αူαိုα်αိုα် αα်αေαြαြα့်αါ။
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. |
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α‘αα်αါαွဲ့α α်းαုံαို αေ့αာαြα်းα‘ားαြα့် $i$ ၏ αα်αွှα်းαα် $4$ ၏ ααိုးαိα်းαြα ်αိုα်း $1$ αှα့် αီαြောα်းαွေ့ααα်။ αို့αြောα့် $i^4$ = $i^8$ = $i^{12}$ = ... = $i^{4n}$ = $1$ αု αုံαေ αှα်αူαိုα်αါαα်။
| Example 1 Simplify each power of $i$. (a) $i^{13}\quad$ (b) $i^{56}\quad$ (c) $i^{-3}$ Solution $\begin{array}{ll} \text{(a)}\quad i^{13} &=i^{12+1}\\ &=i^{12}\cdot i\\ &= (i^{4})^3\cdot i\\ &= (1)^3\cdot i\\ &= i\\ \text{(b)}\quad i^{56} &= (i^{4})^{14}\\ &= 1^{14}\\ &= 1\\ \text{(c)}\quad i^{-3} &= i^{-4+1}\\ &= i^{-4}\cdot i\\ &= (i^{4})^{-1}\cdot i\\ &= (1)^{-1}\cdot i\\ &= 1\cdot i\\ &= i \end{array}$ |
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| 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 $\begin{array}{ll} \text{(a)}\quad -3 i^{5}+2 i^{2} &=-3 i^{4} \cdot i+2(-1) \\ &=-3(1) \cdot i-2 \\ &=-2-3 i \\ \text{(b)}\quad (\sqrt{-32})^3 &=(4 \sqrt{2} \sqrt{-1})^{3} \\ &=(4 \sqrt{2} i)^{3} \\ &=64(2 \sqrt{2}) i^{3}\\ &=128 \sqrt{2} i^{2} \cdot i \\ &=128 \sqrt{2}(-1) i \\ &=-128 \sqrt{2} i \\ \text{(c)}\quad \dfrac{1}{5 i^{3}} &=\dfrac{1}{5} i-3 \\ &=\dfrac{1}{5} i^{-4} \cdot i \\ &=\dfrac{1}{5}\left(i^{4}\right)^{-1} i \\ &=\dfrac{1}{5}(1)^{-1} i \\ &=\dfrac{1}{5} i \end{array}$ |
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Complex Solutions of a Quadratic Equation
Quadratic Equation αα ်αု၏ standard equation form αှာ
- $ax^2+bx+c=0$ αြα ်αြောα်းαိαှိαဲ့αြီး αြα ်αြောα်း
- Quadratic Equation αိုαြေαα်α ေαော solution αို quadratic formula $ \dfrac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}$ αြα့်αှာαူαိုα်αြောα်း
- $b^2-4ac$ αို quadratic function αα ်αု၏ discriminant αု αေါ်αြောα်း
- discriminant < 0 αြα ်αါα αေးαားαော quadratic equation αို αြေαα်α ေαော αိα်းα α ် α‘αြေααှိαြောα်း
αို့αော် discriminant < 0 αြα ်αော quadratic equation αျားα‘αွα် complex solution αို αှာαူαိုα်αါαα်။ α‘ောα်αါ α₯ααာαျားαို αေ့αာαြα့်αါ။
| 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. $\begin{aligned} x &= \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\ &= \dfrac{-5\pm \sqrt{-3}}{2}\\ &= \dfrac{-5}{2}\pm \dfrac{\sqrt{-3}}{2}\\ &= \dfrac{-5}{2}\pm \dfrac{\sqrt{3}\sqrt{-1}}{2}\\ &= \dfrac{-5}{2}\pm \dfrac{\sqrt{3}}{2}i \end{aligned}$ |
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Argand diagram (Complex Plane)
Complex Number αျားαို complex plane αွα် αေααျαα်αှα်αိုα်αြောα်း Part (1) αွα် αα်αြαဲ့αြီး αြα ်αα်။ α‘αိုαါ complex plane αို Argand plane (αို့ααုα်) Gauss Plane αုαα်း αေါ်αα်။ $z=a+bi$ ၏ αα်αေαာαα် Argand plane αွα် $P(a,b)$ αြα ်αα်။ αို့αြောα့်
- $z_1=4+3i$ αα်αေαာαα် Argand plane αွα် $(4,3)$
- $z_2=3-2i$ αα်αေαာαα် Argand plane αွα် $(3,-2)$
- $z_3=-1-5i$ αα်αေαာαα် Argand plane αွα် $(-1,-5)$
- $z_4=-4+3i$ αα်αေαာαα် Argand plane αွα် $(-4,3)$
- $z_5=3i$ αα်αေαာαα် Argand plane αွα် $(0,3)$
- $z_6=-2i$ αα်αေαာαα် Argand plane αွα် $(0,-2)$
- $z_7=-5$ αα်αေαာαα် Argand plane αွα် $(-5,0)$
- $z_8=3$ αα်αေαာαα် Argand plane αွα် $(3,0)$
αို့αြောα့် complex number αα ်αုαို argand diagram (complex plane) αေါ်αှိ coordinate αြα့်αα်း αေါ်αြαေ့αှိαα်။ α₯ααာ
| $\begin{array}{lll} z_1=4+3i & \text{or} & z_1=(4,3)\\ z_2=3-2i & \text{or} & z_2=(3,-2)\\ z_3=-1-5i & \text{or} & z_3=(-1,-5)\\ z_4=-4+3i & \text{or} & z_4=(-4,3)\\ z_5=3i & \text{or} & z_5=(0,3)\\ z_6=-2i & \text{or} & z_6=(0,-2)\\ z_7=-5 & \text{or} & z_7=(-5,0)\\ z_8=3 & \text{or} & z_8=(3,0)\\ \end{array}$ |
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αုαေါ်αြαိုα်αα်။
Modulus and Argument of a Complex Number
Complex number $z =x+yi$ αို Complex plane αွα် plot αုα်αα့်α‘αါ α‘ောα်αါα‘αိုα်း ααα် αိုαါα ို့။
| $\begin{array}{l} \text{By Pythagoras' Theorem}\\ |z|=\sqrt{x^2+y^2}\\ \text{Since}\ \tan{\theta}=\dfrac{y}{x},\\ \theta=\tan^{-1}\dfrac{y}{x}\\ \operatorname{arg}(z)=\tan^{-1}\left(\dfrac{y}{x}\right) \end{array}$ |
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$\operatorname{arg}(z)=\theta$ αို αော်αြαာαွα် $-180^{\circ} < \theta \le 180^{\circ}$ (radian αြα့်αော်αြαော် $-\pi < \theta \le \pi$) interval αြα့် αော်αြαα်။ ၎α်းαို principal argument αုαα်းαေါ်αα်။ αိုαိုαα်αှာ αေးαားαော complex number αα် argand plane ၏ first αှα့် second quadrant αွα်αှိαါα $0<\operatorname{arg}(z)<180^{\circ}$ (positive angle) αြα ်αြီး αေးαားαော complex number αα် argand plane ၏ third αှα့် fourth quadrant αွα်αှိαါα $-180^{\circ}<\operatorname{arg}(z)< 0$ (negative angle) αြα ်αα်။
| $\begin{array}{cccc} \hline z & |z| & \text{location} & \text{argument}\\ \hline a+bi\\(a,b>0) &\sqrt{a^2+b^2} & 1^{\text{st}}\text{quadrant} & \operatorname{arg}(z)=\tan^{-1}\left|\dfrac{b}{a}\right|\\ \hline a+bi\\(a<0,b>0) &\sqrt{a^2+b^2} & 2^{\text{nd}} \text{quadrant} & \operatorname{arg}(z)=\left(180^{\circ}-\tan^{-1}\left|\dfrac{b}{a}\right|\right)\\ \hline a+bi \\(a,b<0) &\sqrt{a^2+b^2} & 3^{\text{rd}} \text{quadrant} & \operatorname{arg}(z)=-\left(180^{\circ}-\tan^{-1}\left|\dfrac{b}{a}\right|\right)\\ \hline a+bi\\(a>0,b<0) & \sqrt{a^2+b^2}& 4^{\text{th}} \text{quadrant} &\operatorname{arg}(z)=-\tan^{-1}\left|\dfrac{b}{a}\right|\\ \hline a \\(a>0) & |a| & \text{on positive}\ x\text{-axis} & 0^{\circ}\\ \hline a \\(a<0) & |a| & \text{on negative}\ x\text{-axis} & 180^{\circ}\\ \hline bi\\(b>0) & |b| & \text{on positive}\ y\text{-axis} & 90^{\circ}\\ \hline bi \\(b<0)& |b| &\text{on negative}\ y\text{-axis} & -90^{\circ}\\ \hline \end{array}$ |
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| 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) $\quad z=3-5i$ $\quad\quad |z|=\sqrt{(3)^2+(-5)^2}=\sqrt{34}$ $\quad\quad \tan^{-1}\left(\dfrac{5}{3}\right)=59.04^{\circ}$ $\therefore\quad \operatorname{arg}(z)=-59.04^{\circ}$ |
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$z=x+yi$ αြα ်αα်αိုαါα ို့။ αိုα‘αါ $\overline{z}=x-yi$ αြα ်αα်။ αို့αြောα့်
αြα ်αα်။ αို့αြောα့် αα်αα့် complex number α‘αွα်ααို α‘ောα်αါα‘αို αα် α‘αြဲαှα်αα်αα်။
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 $\begin{aligned} \text { Let } w &=\dfrac{z+i}{z-i} \\\\ \overline{w} &= \overline{\left(\dfrac{z+i}{z-i}\right)} \\\\ &=\dfrac{\overline{z+i}}{\overline{z-i}} \\\\ &=\dfrac{\overline{z}+\overline{i}}{\overline{z}-\overline{i}} \\\\ &=\dfrac{\overline{z}-i}{\overline{z}+i}\\\\ \therefore\ w-\overline{w} &=\dfrac{z+i}{z-i}-\dfrac{\overline{z}-i}{\overline{z}+i} \\\\ &=\dfrac{(z+i)(\overline{z}+i)-(z-i)(\overline{z}-i)}{(z-i)(\overline{z}+i)} \\\\ &=\dfrac{z \overline{z}+z i+\overline{z} i-1-z \overline{z}+z i+\overline{z}^{i}+1}{(z-i)(\overline{z}+i)} \\\\ &=\dfrac{2(z+\overline{z}) i}{(z-i)(\overline{z}+i)}\\\\ \dfrac{w-\overline{w}}{2 i}&=\dfrac{z+\overline{z}}{(z-i)(\overline{z}+i)} \\\\ \operatorname{Im}(w)&=\dfrac{z+\overline{z}}{(z-i)(\overline{z}+i)} \\\\ \text { By the}\ & \text{problem, } \\\\ \text { Im }(w)&=0 \\\\ \dfrac{z+\overline{z}}{(z-i)(\overline{z}+i)}&=0\\\\ \therefore \quad z+\overline{z}&=0 \\\\ \quad\quad \dfrac{z+\overline{z}}{2}&=0 \\\\ \therefore\quad \operatorname{Re}(z)&=0 \end{aligned}$ |
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| Example 6 Given that $z \in \mathbb{C}$ and $|z|=1$, show that $\operatorname{Re}\left(\dfrac{z-1}{z+1}\right)=0$. Solution $\begin{aligned} \text { Let } &\frac{z-1}{z+1}=w \\\\ \therefore \overline{w} &=\overline{\left(\frac{z-1}{z+1}\right)} \\\\ &=\frac{\overline{z-1}}{\overline{z+1}} \\\\ &=\frac{\overline{z}-1}{\overline{z}+1} \\\\ w+\overline{w}&=\frac{z-1}{z+1}+\frac{\overline{z}-1}{\overline{z}+1}\\\\ w+\overline{w} &=\dfrac{z \overline{z}+z-\overline{z}-1+z \overline{z}-z+\overline{z}-1}{(z+1)(\overline{z}+1)} \\\\ w+\overline{w} &=\dfrac{2(z \overline{z}-1)}{(z+1)(\overline{z}+1)} \\\\ \therefore \quad \dfrac{w+\overline{w}}{2} &=\dfrac{z \overline{z}-1}{(z+1)(\overline{z}+1)} \\\\ \operatorname{Re}(w) &=\dfrac{|z|^{2}-1}{(z+1)(\overline{z}+1)}\\\\ &=0 \quad(\because|z|=1) \\\\ \therefore \operatorname{Re}\left(\dfrac{z-1}{z+1}\right)&=0 . \end{aligned}$ |
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Example 7
$\begin{aligned} 1.\quad\quad\quad\quad\quad\quad z & =x+y i \\\\ z^{2} & =15+8 i \\\\ (x+y i)^{2} & =15+8 i \\\\ x^{2}+2 x y i+y^{2} i^{2}&=15+8 i \\\\ \therefore \quad x^{2}-y^{2}& =15\quad \ldots(1) \\\\ \quad\quad 2 x y &=8 \quad \ldots(2)\\\\ \therefore x y=4 & \Rightarrow y=\dfrac{4}{x} \\\\ \therefore x^{2}-\left(\dfrac{16}{x^{2}}\right)&=15 \\\\ x^{4}-16&=15 x^{2} \\\\ x^{4}-15 x^{2}-16&=0 \\\\ \left(x^{2}-16\right)\left(x^{2}+1\right)&=0 \\\\ x^{2}=16\ \text { or }\ & x^{2}=-1 \\\\ \therefore x=\pm 4\ \text { or }\ & x=i \end{aligned}$ $\text { Since }\ x \text { is an integer, } x=\pm 4$ When $x=4, y=\dfrac{4}{4}=1$ When $x=-4, y=\dfrac{4}{-4}=-1$ $\quad z=4+i \text { (or) } z=-(4+i)$ $\begin{aligned} \therefore \quad \sqrt{15+8 i} &=\sqrt{z^{2}} \\\\ &=z \\\\ &=\pm(4+i) \end{aligned}$ $\begin{array}{l} \text { 2. } w^{2}-(2+3 i) w-5+i=0 \\\\ \text { Comparing with } a w^{2}+b w+c =0, \\\\ \quad a=1, b=-(2+3 i) \text { and } c=-5+i \\\\ \therefore\quad w =\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\\\ \quad\quad\quad =\dfrac{2+3 i \pm \sqrt{(2+3 i)^{2}-4(-5+i)}}{2} \\\\ \quad\quad\quad =\dfrac{2+3 i \pm \sqrt{4+12 i-9+20-4 i}}{2}\\\\ \quad\quad\quad =\dfrac{2+3 i \pm \sqrt{15+8 i}}{2} \\\\ \quad\quad\quad =\dfrac{2+3 i \pm(4+i)}{2} \\\\ \therefore \quad w=\dfrac{2+3 i+4+i}{2}=3+2 i \\\\ \quad\quad\quad\quad(\text { or }) \\\\ \quad\quad w=\dfrac{2+3 i-4-i}{2}=-1+i \end{array}$ |
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| 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 where $z_1$ is the impedance (in ohms) of pathway 1 and $z_2$ is the impedance (in ohms) of pathway 2.
$\begin{aligned} z_{1}&=9+16 i \text { ohm } \\\\ z_{2}&=20-10 i \text { ohm } \\\\ \therefore\quad z_{1}+z_{2}&=29+6 i \text { ohm } \\\\ z_{1} \cdot z_{2}& =340+230 i \text { ohm } \\\\ \dfrac{1}{z}&=\dfrac{1}{z_{1}}+\dfrac{1}{z_{2}} \\\\ &=\dfrac{z_{1}+z_{2}}{z_{1} z_{2}}\\\\ z &=\dfrac{z_{1} z_{2}}{z_{1}+z_{2}} \\\\ &=\dfrac{340+230 i}{29+6 i} \times \dfrac{29-6 i}{29-6 i} \\\\ &=\dfrac{11240+4630 i}{877} \\\\ &=\dfrac{11240}{877}+\dfrac{4630}{877} i \text { ohm } \end{aligned}$ |
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Exercises
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