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- Question 5, Exercise 1.4
- tan. =====Question 5===== If $\cos \alpha+\cos \beta+\cos \gamma=\sin \alpha+\sin \beta+\sin \gamma=0$, show that: (i) $\cos 3 \alpha+\cos 3 \beta+\cos 3 \gamma=3 \cos (\alpha+\beta+\gamma)$. (ii) $\sin 3 \alpha+\sin 3 \beta+\sin 3 \gamma=3 \sin
- Question 4, Exercise 1.3
- 3 ; 2 z-(2+5 i) \omega=2+3 i$. ** Solution. ** \begin{align} &(1-i) z+(1+i) \omega=3 \quad \cdots(1)... cdots(2) \end{align} Multiplying Eq. (1) by $2$: \begin{align} &(2-2i)z+(2+2i) \omega=6 \quad \cdots (3) \end{align} Multiplying Eq. (2) by $(1-i)$: \begin{align} &2(1-i)z-(1-i) (2+5 i)\omega=(1-i) (2+3... i \quad \cdots (4) \end{align} $(3)-(4)$ implies \begin{align} (9+5i) \omega=1-i \end{align} \begin{al
- Question 10, Exercise 1.2
- left|-\overline{z_{!}}\right|.$$ **Solution.** \begin{align} |z_1| &= \sqrt{(-3)^2 + (2)^2} \\ &= \... + 4} = \sqrt{13} \,\, -- (1) \end{align} Now \begin{align} -z_1 &= -(-3 + 2i) = 3 - 2i\\ \implies ... + 4} = \sqrt{13} \,\, -- (2) \end{align} Also \begin{align} \overline{z_1} &= -3 - 2i \\ \implies ... + 4} = \sqrt{13} \,\, -- (3) \end{align} Now \begin{align} -\overline{z_1} &= -(-3 - 2i) = 3 + 2i
- Question 1, Exercise 1.3
- o linear functions: $z^{2}+169$. **Solution.** \begin{align} & z^{2} + 169 \\ = & z^{2} - (13i)^2 \... linear functions: $2 z^{2}+18$. **Solution.** \begin{align} & 2z^2 + 18 \\ = &2(z^2 - (3i)^2)\\ = ... near functions: $3 z^{2}+363$. **Solution.** \begin{align} & 3z^2 + 363 \\ = & 3(z^2 - (11i)^2)\\... unctions: $z^{2}+\dfrac{3}{25}$. **Solution.** \begin{align} & z^2 + \dfrac{3}{25} \\ = & z^2 - \lef
- Question 9, Exercise 1.2
- +4 i)^{-1}$. **Solution.** Suppose $z=2+4i$. \begin{align} Re(2+4i)^{-1} & = Re(z^{-1}) = \dfrac{R... \dfrac{2}{20}\\ &= \dfrac{1}{10}. \end{align} \begin{align} Im(2+4i)^{-1} & = Im(z^{-1}) = -\dfrac{... |z|^4}. \] First, note $Re(z)=3$, $Im(z)=-2$ and \begin{align} |z| &= \sqrt{3^2 + (-2)^2} \\&= \sqrt{1... gn} Then \[|z|^4=169.\] Using in above formulas \begin{align} Re((3 - 2i)^{-2}) &= \frac{(3)^2 - (-2)
- Question 1, Exercise 1.4
- tion. ** Let $z=x+iy=2 + i 2 \sqrt{3}$. We have \begin{align} r & = \sqrt{x^2 + y^2} = \sqrt{2^2 + (... = \sqrt{4 + 12} = \sqrt{16} = 4. \end{align} and \begin{align} \alpha & = \tan^{-1}\left|\frac{y}{x}\r... orm. ** Solution. ** Let $z=x+iy=3-i \sqrt{3}$ \begin{align} r &= \sqrt{x^2 + y^2} = \sqrt{3^2 + (-\... 3} = \sqrt{12} = 2\sqrt{3}. \end{align} Next, \begin{align} \alpha &= \tan^{-1}\left|\frac{y}{x}\ri
- Question 7, Exercise 1.4
- {\pi}{4}$ ** Solution. ** Suppose $z=x+iy$, as \begin{align*} &\arg (z-1)=-\dfrac{\pi}{4} \\ \implie... n. ** Suppose $z=x+iy$, then $\bar{z}=x-iy$. As \begin{align*} &z \bar{z}=4\left|e^{i \theta}\right| ... arg (z-4) \leq \dfrac{\pi}{3}$ ** Solution. ** \begin{align*} &-\frac{\pi}{3} \leq \arg (z-4) \leq \... right) \leq \dfrac{\pi}{6}$ ** Solution. ** \begin{align*} &0 \leq \arg \left(\frac{z-4}{1+i}\rig
- Question 4, Exercise 1.1
- ollowing: $(2+3i)x+(1+3i)y+2=0$ **Solution.** \begin{align}&(2+3i)x+(1+3i)y+2=0\\ \implies &(2x+y+2... 0.\end{align} Comparing real and imaginary parts \begin{align} 2x+y+2&=0 \quad \cdots(1)\\ 3x+3y&=0\quad \cdots (2) \end{align} From (2), \begin{align} &3x=-3y \\ x=-y \quad ... (3) \end{align} Putting value of $x$ in (1) \begin{align}2(-y)+y+2&=0\\ -2y+y&=-2\\ -y&=-2\\ y&=2
- Question 8, Exercise 1.2
- ** Given: $$|2z-i|=4.$$ Put $z=x+i y$, we have \begin{align} & |2(x+iy)-i|=4 \\ \implies & |2x+i(2y-... 2+(2y-1)^2}=4 \end{align} Squaring on both sides \begin{align} & (2x)^2+(2y-1)^2 = 16\\ \implies & 4x^... en: $$|z-1|=|\bar{z}+i|.$$ Put $z=x+iy$, we have \begin{align} & |(x+iy)-1| = |(x-iy)+i| \\ \implies &... (y-1)^2} \end{align} Squaring both sides, we get \begin{align} & (x-1)^2 + y^2 = x^2 + (y-1)^2 \\ \imp
- Question 2, Exercise 1.3
- pleting square: $z^{2}-6 z+2=0$. **Solution.** \begin{align} & z^2 - 6z + 2 = 0 \\ \implies & z^2 - ... \end{align} Take the square root of both sides: \begin{align} &z - 3 = \pm \sqrt{7} \\ \implies &z =... : $-\dfrac{1}{2} z^{2}-5 z+2=0$. **Solution.** \begin{align} -\dfrac{1}{2} z^{2} - 5z + 2& = 0 \end{... tiply through by $-2$ to eliminate the fraction: \begin{align} z^2 + 10z - 4 &= 0 \\ z^2 + 10z +25-25
- Question 6(i-ix), Exercise 1.4
- i \sin 315^{\circ}\right)$ ** Solution. ** \begin{align} &\sqrt{2}\left(\cos 315^{\circ}+i \sin ... rc}+i \sin 210^{\circ}\right)$ ** Solution. ** \begin{align*} &5\left(\cos 210^\circ + i \sin 210^\c... sin \dfrac{3 \pi}{2}\right)$ ** Solution. ** \begin{align*} &2\left(\cos \frac{3\pi}{2} + i \sin \... in \dfrac{5 \pi}{6}\right)$ ** Solution. ** \begin{align*} &4\left(\cos \frac{5\pi}{6} + i \sin \
- Question 2, Exercise 1.1
- form $x+iy$: $(3+2i)+(2+4i)$ ** Solution. ** \begin{align}&(3+i2)+(2+i4)\\ =&(3+2)+(i2+i4)\\ =&5+i... he form $x+iy$: $(4+3i)-(2+5i)$ **Solution.** \begin{align}&(4+3i)-(2+5i)\\ =&(4-2)+(3i-5i)\\ =&2-2... he form $x+iy$: $(4+7i)+(4-7i)$ **Solution.** \begin{align} &(4+7i)+(4-7i)\\ =&(4+4)+(7i-7i)\\ =&8+... the form $x+iy$: $(2+5i)-(2-5i)$ **Solution.** \begin{align} &(2+5i)-(2-5i)\\ =&(2-2)+(5i+5i)\\ =&0+
- Question 1, Exercise 1.1
- 1(i)==== Evaluate ${{i}^{31}}$. **Solution.** \begin{align}{{i}^{31}}&=i\cdot{{i}^{30}}\\ &=i\cdot{... {15}}\\ &=i\cdot{{\left( -1 \right)}^{15}} \quad \because i^2=-1\\ &=i\cdot(-1)\\ &=-i.\end{align} GO... ate ${{\left( -i \right)}^{6}}$. **Solution.** \begin{align} {{\left( -i \right)}^{23}}&=(-1)^{23} i... ( -1 \right)}^{\frac{-13}{2}}}$. **Solution.** \begin{align}{{\left( -1 \right)}^{\frac{-23}{2}}}&={
- Question 7, Exercise 1.1
- 2 i$. **Solution.** Suppose $$z=11+12i$$ Then \begin{align}|z|&= \sqrt{(11)^2+(12)^2}\\ &=\sqrt{265... . **Solution.** Suppose $z=(2+3i)ā(2+6i)$, then \begin{align}z&=2+3iā2ā6i\\ &=-3i \end{align} Now \begin{align} |z| &= \sqrt{0^2+(-3)^2} \\ &= \sqrt{9} ... **Solution.** Suppose $$z=(2-i)(6+3 i),$$ then \begin{align} |z|&=|(2-i)(6+3 i)|\\ &=|(2-i)| |(6+3
- Question 3, Exercise 1.2
- maginary part of $z$ is zero. i.e. $b=0$. Then \begin{align} &z=a \\ \implies &\bar{z}=a \end{align}... that $\overline{z}=z$, then we $z$ is real. As \begin{align}& z=\bar{z}\\ \Rightarrow \quad & a+ib=a... row \quad & 2ib=0\\ \Rightarrow \quad & b=0\quad \because \quad 2i\neq 0\end{align} Then (1) becomes $$z=a+i(0)=a$$ This gives $z$ is real. ====Question 3