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- Question 10, Exercise 1.2
- .** Given \[z_1 = -3 + 2i, \quad z_2 = 1 - 3i\] Then \begin{align} \frac{z_1}{z_2} &= \frac{-3 + 2i}{1... - 2i, \quad \overline{z_2} = 1 + 3i. \end{align} Then \begin{align} \frac{\overline{z_1}}{\overline{z_2... \\ &= -3 + 11i + 6\\ &= 3 + 11i. \end{align} Then \[ \overline{z_1 z_2}= 3 - 11i. -- (i)\] Now \[ ... z_1} = -3 - 2i, \quad \overline{z_2} = 1 + 3i. \] Then \begin{align} \overline{z_1}\,\, \overline{z_2}
- Question 3, Exercise 1.2
- bb{R}\, ... (1)$$ First suppose that $z$ is real, then we shall prove $\overline{z}=z$. Since $z$ is real, imaginary part of $z$ is zero. i.e. $b=0$. Then \begin{align} &z=a \\ \implies &\bar{z}=a \end{a... }$. Now conversly suppose that $\overline{z}=z$, then we $z$ is real. As \begin{align}& z=\bar{z}\\ \R... uad & b=0\quad \because \quad 2i\neq 0\end{align} Then (1) becomes $$z=a+i(0)=a$$ This gives $z$ is real
- Question 7, Exercise 1.1
- $11+12 i$. **Solution.** Suppose $$z=11+12i$$ Then \begin{align}|z|&= \sqrt{(11)^2+(12)^2}\\ &=\sqrt... +6 i)$. **Solution.** Suppose $z=(2+3i)ā(2+6i)$, then \begin{align}z&=2+3iā2ā6i\\ &=-3i \end{align} Now... i)$. **Solution.** Suppose $$z=(2-i)(6+3 i),$$ then \begin{align} |z|&=|(2-i)(6+3 i)|\\ &=|(2-i)| |(... **Solution.** Suppose $$z=\dfrac{3-2 i}{2+i},$$ then \begin{align} |z|&=\left|\dfrac{3-2 i}{2+i} \righ
- Question 1, Exercise 1.3
- {2})=z+\dfrac{3}{2}$ is the factor of polynomial. Then by using synthetic division: \begin{align} \begin... 0 $$ So $z-(1)=z-1$ is the factor of polynomial. Then by using synthetic division: \begin{align} \begin... ign} So $z-(-3)=z+3$ is the factor of polynomial. Then by using synthetic division: Now, by synthetic di... nd{align} So $z-2$ is the factor of polynomial.\\ Then by using synthetic division: \begin{align} \begin
- Question 3, Exercise 1.3
- 1,\quad b = 6,\quad \text{and}\quad c = -48.$$ Then \begin{align} z& = \dfrac{{-6 \pm \sqrt{36-4(1)(... $ a = 1, \quad b = -\dfrac{1}{2}, \quad c = 17 $$ Then \begin{align} z &= \dfrac{{-\left(-\dfrac{1}{2}\r... $$ Where $$ a = 1, \quad b = -6, \quad c = 25 $$ Then \begin{align} z &= \dfrac{{-(-6) \pm \sqrt{{(-6)^... $$ Where $$ a = 1, \quad b = -9, \quad c = 11 $$ Then \begin{align} z &= \dfrac{{-(-9) \pm \sqrt{{(-9)^
- Question 1, Review Exercise
- 0,0)$</collapse> iv. If $z$ is a complex number then its mirror image is * (a) $|z|$ * (b) ... collapse> vi. If $z_{1}=3+2 i$ and $z_{2}=5+6 i$ then * (a) $z_{1}>z_{2}$ * (b) $z_{1}<z_{... %: argand</collapse> viii. If $\mathrm{z}=3+4 i$ then $\mathrm{z}^{-1}$ is * (a) $\left(\frac{1}{3... pse> x. If $\left(\frac{1+i}{1-i}\right)^{n}=1$ then least positive value of $n$ is * (a) $1$
- Question 6, Exercise 1.1
- number $4-3 i$. **Solution.** Given: $z=4-3 i$, then $\bar{z}=4+3i$. ====Question 6(ii)==== Find the... ac{-1}{5}}\\ =&2+\sqrt{\dfrac{1}{5}}i,\end{align} Then $$\bar{z}=2-\sqrt{\dfrac{1}{5}}i$$ ====Question ... tion.** Given: $z=\dfrac{5 }{2}i-\dfrac{7}{8},$ then $\bar{z}=-\dfrac{5 }{2}i-\dfrac{7}{8}$. GOOD ==
- Question 9, Exercise 1.2
- = \sqrt{3^2 + (-2)^2} \\&= \sqrt{13}. \end{align} Then \[|z|^4=169.\] Using in above formulas \begin{al... _2^2\): \[x_2^2 - y_2^2 = 2^2 - 5^2 = -21.\] Then, compute \(x_1^2 - y_1^2\): \[x_1^2 - y_1^2 = 4... _2 + y_1 y_2 = 3 \cdot 2 + (-7) \cdot 5 = -29\] Then, compute \(x_2 y_1 - x_1 y_2\): \[x_2 y_1 - x_1
- Question 7, Exercise 1.4
- heta}\right|$ ** Solution. ** Suppose $z=x+iy$, then $\bar{z}=x-iy$. As \begin{align*} &z \bar{z}=4\le... 2} \arg (z+i)$ ** Solution. ** Assume $z=x+iy$, then \begin{align*} &\dfrac{1}{2} \arg (z-i)=\dfrac{\p
- Question 5, Exercise 1.1
- ac{1-18i}{2-i}$ **Solution.** Suppose $z=x+iy$, then $\bar{z}=x-iy$. So we have \begin{align}&4z-3\bar
- Question 4, Exercise 1.2
- 2}\right|$. **Solution.** Given $$z_{1}=2-3i$$ Then \begin{align}|z_1|&=\sqrt{(2)^2+(-3)^2}\\ &=\sqrt
- Question 5, Exercise 1.2
- == If $z_1$ and $z_2$ are two any complex numbers then prove that $|z_1+z_2|^2-|z_1-z_2|^2=4Re(z_1)Re(z_
- Question 8, Exercise 1.2
- rac{1}{2} Re(i \bar{z}) = 4.$$ Put $z = x + i y$, then $\bar{z} = x - i y$. We have \begin{align} & \df
- Question 1, Exercise 1.4
- . ** Solution. ** Let \(z=x+iy= i - 1=-1+i \). Then \begin{align} r & = \sqrt{x^2 + y^2} = \sqrt{(-1)
- Question 2, Exercise 1.4
- pi}{3}+i \sin \dfrac{\pi}{3}=e^{i\frac{\pi}{3}}$. Then \begin{align} z_1 z_2 & = e^{i\frac{\pi}{6}} \cdo