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- Question 1, Exercise 1.3
- polynomial into linear functions: $z^{2}+169$. **Solution.** \begin{align} & z^{2} + 169 \\ = & z^{2} - (... olynomial into linear functions: $2 z^{2}+18$. **Solution.** \begin{align} & 2z^2 + 18 \\ = &2(z^2 - (3i)^... nomial into linear functions: $3 z^{2}+363$. **Solution.** \begin{align} & 3z^2 + 363 \\ = & 3(z^2 - (1... into linear functions: $z^{2}+\dfrac{3}{25}$. **Solution.** \begin{align} & z^2 + \dfrac{3}{25} \\ = & z^
- Question 6(i-ix), Exercise 1.4
- os 315^{\circ}+i \sin 315^{\circ}\right)$ ** Solution. ** \begin{align} &\sqrt{2}\left(\cos 315^{\circ... t(\cos 210^{\circ}+i \sin 210^{\circ}\right)$ ** Solution. ** \begin{align*} &5\left(\cos 210^\circ + i \s... c{3 \pi}{2}+i \sin \dfrac{3 \pi}{2}\right)$ ** Solution. ** \begin{align*} &2\left(\cos \frac{3\pi}{2} +... {5 \pi}{6}+i \sin \dfrac{5 \pi}{6}\right)$ ** Solution. ** \begin{align*} &4\left(\cos \frac{5\pi}{6} +
- Question 2, Exercise 1.1
- ex number in the form $x+iy$: $(3+2i)+(2+4i)$ ** Solution. ** \begin{align}&(3+i2)+(2+i4)\\ =&(3+2)+(i2+i... lex number in the form $x+iy$: $(4+3i)-(2+5i)$ **Solution.** \begin{align}&(4+3i)-(2+5i)\\ =&(4-2)+(3i-5i... lex number in the form $x+iy$: $(4+7i)+(4-7i)$ **Solution.** \begin{align} &(4+7i)+(4-7i)\\ =&(4+4)+(7i-7... lex number in the form $x+iy$: $(2+5i)-(2-5i)$ **Solution.** \begin{align} &(2+5i)-(2-5i)\\ =&(2-2)+(5i+5i
- Question 3, Exercise 1.3
- ratic equation: $\dfrac{1}{3} z^{2}+2 z-16=0$. **Solution.** Given \begin{align}&\dfrac{1}{3}z^{2}+2 z-16=0... 57}}}{2} \\ &= -3 \pm \sqrt{57} \end{align} Hence Solution set $=\{ -3 \pm \sqrt{57} \}$. ====Question 3(i... uadratic equation: $z^{2}-\frac{1}{2} z+17=0$. **Solution.** Given $$ z^{2} - \frac{1}{2}z + 17 = 0 $$ Usi... 1 \pm \sqrt{271}i}{4} \end{align} Therefore, the solution set is: $\left\{\dfrac{1 \pm \sqrt{271}i}{4}\righ
- Question 6(x-xvii), Exercise 1.4
- rac{5 \pi}{4}+i \sin \dfrac{5 \pi}{4}\right)$ ** Solution. ** //Do yourself as previous parts.// =====Ques... rac{7 \pi}{4}+i \sin \dfrac{7 \pi}{4}\right)$ ** Solution. ** //Do yourself as previous parts.// =====Q... \dfrac{5\pi}{2}+i \sin \dfrac{5\pi}{2}\right)$ ** Solution. ** //Do yourself as previous parts.// =====Ques... \dfrac{\pi}{4}+i \sin \dfrac{\pi}{4}\right)$ ** Solution. ** //Do yourself as previous parts.// =====Que
- Question 2, Exercise 1.3
- quation by completing square: $z^{2}-6 z+2=0$. **Solution.** \begin{align} & z^2 - 6z + 2 = 0 \\ \implies ... pleting square: $-\dfrac{1}{2} z^{2}-5 z+2=0$. **Solution.** \begin{align} -\dfrac{1}{2} z^{2} - 5z + 2& =... \implies z &= -5 \pm \sqrt{29} \end{align} Hence solution set$=\{-5 \pm \sqrt{29}\}$ ====Question 2(iii)=... uation by completing square: $4 z^{2}+5 z=14$. **Solution.** \begin{align} 4z^{2} + 5z &= 14\\ z^{2} + \d
- Question 8, Exercise 1.2
- $ in terms of $x$ and $y$ by taking $z=x+i y$. **Solution.** Given: $$|2z-i|=4.$$ Put $z=x+i y$, we have ... $ in terms of $x$ and $y$ by taking $z=x+i y$. **Solution.** Given: $$|z-1|=|\bar{z}+i|.$$ Put $z=x+iy$, w... $ in terms of $x$ and $y$ by taking $z=x+i y$. **Solution.** Given: $$|z-4i| + |z+4i| = 10.$$ Put $z = x ... $ in terms of $x$ and $y$ by taking $z=x+i y$. **Solution.** Given: $$\dfrac{1}{2} Re(i \bar{z}) = 4.$$ P
- Question 9, Exercise 1.2
- nd real and imaginary parts of $(2+4 i)^{-1}$. **Solution.** Suppose $z=2+4i$. \begin{align} Re(2+4i)^{-... and imaginary parts of $(3-\sqrt{-4})^{-2}$. **Solution.** Suppose $z=3 - \sqrt{-4}=3-2i$. We will us... ts of $\left(\dfrac{7+2 i}{3-i}\right)^{-1}$. **Solution.** We use the following formulas: \[Re\left(\... of $\left(\dfrac{4+2 i}{2+5 i}\right)^{-2}$. **Solution.** We will use the following formulas: \begin
- Question 10, Exercise 1.2
- _{!}}\right|=\left|-\overline{z_{!}}\right|.$$ **Solution.** \begin{align} |z_1| &= \sqrt{(-3)^2 + (2)^2} ... )}=\frac{\overline{z_{1}}}{\overline{z_{2}}}$. **Solution.** Given \[z_1 = -3 + 2i, \quad z_2 = 1 - 3i\] ... z_{2}}=\overline{z_{1}}\,\, \overline{z_{2}}$. **Solution.** Given \[ z_1 = -3 + 2i, \quad z_2 = 1 - 3i.... {1}+z_{2}}=\overline{z_{1}}+\overline{z_{2}}$. **Solution.** Given \[ z_1 = -3 + 2i, \quad z_2 = 1 - 3i.
- Question 7, Exercise 1.4
- Cartesian form: $\arg (z-1)=-\dfrac{\pi}{4}$ ** Solution. ** Suppose $z=x+iy$, as \begin{align*} &\arg (z... form: $z \bar{z}=4\left|e^{i \theta}\right|$ ** Solution. ** Suppose $z=x+iy$, then $\bar{z}=x-iy$. As \b... {\pi}{3} \leq \arg (z-4) \leq \dfrac{\pi}{3}$ ** Solution. ** \begin{align*} &-\frac{\pi}{3} \leq \arg (z-... dfrac{z-4}{1+i}\right) \leq \dfrac{\pi}{6}$ ** Solution. ** \begin{align*} &0 \leq \arg \left(\frac{z-4
- Question 1, Exercise 1.1
- ====Question 1(i)==== Evaluate ${{i}^{31}}$. **Solution.** \begin{align}{{i}^{31}}&=i\cdot{{i}^{30}}\\ &... (ii)==== Evaulate ${{\left( -i \right)}^{6}}$. **Solution.** \begin{align} {{\left( -i \right)}^{23}}&=(-1... luate ${{\left( -1 \right)}^{\frac{-13}{2}}}$. **Solution.** \begin{align}{{\left( -1 \right)}^{\frac{-23}... Evaluate $\dfrac{2}{(-1)^{\frac{3}{2}}}$. GOOD **Solution.** \begin{align}{{\left( -1 \right)}^{\frac{15}{
- Question 3, Exercise 1.1
- plify the following $\dfrac{(2+i)(3-2i)}{1+i}$ **Solution.** \begin{align}&\dfrac{(2+i)(3-2i)}{1+i}\\ =&\df... Simplify the following $\dfrac{1+i}{(2+i)^2}$ **Solution.** \begin{align}&\dfrac{1+i}{(2+i)^2}\\ =&\dfrac... the following $\dfrac{1}{3+i}-\dfrac{1}{3-i}$ **Solution.** \begin{align}&\dfrac{1}{3+i}-\dfrac{1}{3-i}\\... Simplify the following $(1+i)^{-2}+(1-i)^{-2}$ **Solution.** \begin{align}&(1+i)^{-2}+(1-i)^{-2}\\ =&\dfra
- Question 7, Exercise 1.1
- 7(i)==== Find the magnitude of the $11+12 i$. **Solution.** Suppose $$z=11+12i$$ Then \begin{align}|z|&=... = Find the magnitude of the $(2+3 i)-(2+6 i)$. **Solution.** Suppose $z=(2+3i)−(2+6i)$, then \begin{align}z... ==== Find the magnitude of the $(2-i)(6+3 i)$. **Solution.** Suppose $$z=(2-i)(6+3 i),$$ then \begin{align... ind the magnitude of the $\dfrac{3-2 i}{2+i}$. **Solution.** Suppose $$z=\dfrac{3-2 i}{2+i},$$ then \begin
- Question 4, Exercise 1.1
- n each of the following: $(2+3i)x+(1+3i)y+2=0$ **Solution.** \begin{align}&(2+3i)x+(1+3i)y+2=0\\ \implies... ollowing: $\dfrac{x}{(1+i)}+\dfrac{y}{1-2i}=1$ **Solution.** \begin{align}&\dfrac{x}{(1+i)}+\dfrac{y}{1-2... x}{(2+i)}=\dfrac{1-5i}{(3-2i)}+\dfrac{y}{2-i}$ **Solution.** \begin{align}&\dfrac{x}{(2+i)}=\dfrac{(1-5i)... ch of the following: $x(1+i)^2+y(2-i)^2=3+10i$ **Solution.** \begin{align}&x(1+i)^2+y(2-i)^2=3+10i\\ \Rig
- Question 6, Exercise 1.1
- d the conjugate of the complex number $4-3 i$. **Solution.** Given: $z=4-3 i$, then $\bar{z}=4+3i$. ====... d the conjugate of the complex number $3 i+8$. **Solution.** Do Yourself ====Question 6(iii)==== Find th... f the complex number $2+\sqrt{\dfrac{-1}{5}}$. **Solution.** Given: \begin{align}z=&2+\sqrt{\dfrac{-1}{5}... complex number $\dfrac{5 }{2}i-\dfrac{7}{8}$. **Solution.** Given: $z=\dfrac{5 }{2}i-\dfrac{7}{8},$ then