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- Question 4, Exercise 1.3
- {53}i\end{align} Put value of $\omega$ in (1), we have \begin{align} &(1-i) z+(1+i)\left(\dfrac{2}{53}-\... c{155}{106}+\dfrac{145}{106}i\end{align} Thus, we have $$z=\dfrac{155}{106}+\dfrac{145}{106}i, \omega=\d... -12+10i \quad \cdots(4) \end{align} $(3)-(4)$,we have \begin{align} &(-1-8i+4-6i)\omega=3-i+12-10i\\ \i... c{36}{205}+\dfrac{373}{205}i \end{align} Thus, we have: $$z = \dfrac{36}{205} + \dfrac{373}{205}i;\quad
- Question 10, Exercise 1.2
- -- (4) \end{align} From (1), (2), (3) and (4), we have: $$\left| z_1 \right| = \left| -z_1 \right| = \le... + \frac{7}{10}i. \,\, -- (i) \end{align} Now, we have \begin{align} \overline{z_1} = -3 - 2i, \quad \ov... frac{7}{10}i.\,\, -- (ii)$$ From (i) and (ii), we have \[ \overline{\left( \frac{z_1}{z_2} \right)} = \f... 11i. -- (ii) \end{align} From (i) and (ii), we have \[ \overline{z_1 z_2} = \overline{z_1} \overline
- Question 8, Exercise 1.2
- ution.** Given: $$|2z-i|=4.$$ Put $z=x+i y$, we have \begin{align} & |2(x+iy)-i|=4 \\ \implies & |2x+i... * Given: $$|z-1|=|\bar{z}+i|.$$ Put $z=x+iy$, we have \begin{align} & |(x+iy)-1| = |(x-iy)+i| \\ \impli... n: $$|z-4i| + |z+4i| = 10.$$ Put $z = x + iy$, we have \begin{align} & |(x + iy) - 4i| + |(x + iy) + 4i|... Put $z = x + i y$, then $\bar{z} = x - i y$. We have \begin{align} & \dfrac{1}{2}Re(i(x-iy)) = 4 \\ \i
- Question 3, Exercise 1.4
- d z=|z|e^{i\theta} \,\,-- (1) \end{align*} Now we have given \begin{align*} & \left(x_{1}+i y_{1}\right)... \cdots z_n = z. \end{align*} By using $(1)$, we have \begin{align*} &|z_1| e^{i\theta_1}\cdot |z_2| e^... right)=a^{2}+b^{2}\,\,-- (4)$$ Now from $(3)$, we have $$ \sum_{r=1}^{n} \theta_r = \theta + 2k\pi, \qua... results. **Alternative Method for Part (i)** We have given \begin{align*} & \left(x_{1}+i y_{1}\right)
- Question 2, Exercise 1.2
- plicative assocative law} \end{align} That is, we have proved $$(z_1 z_2)(z_3 z_4)=(z_1 z_3) (z_2 z_4) .... licative associative law} \end{align} That is, we have proved $$(z_1 z_3) (z_2 z_4)=z_3 (z_1 z_2) z_4 ... (ii)$$ From (i) and (ii), we have the required result. **Remark:** For any three complex numbers $z_1$, $z_2$ and $z_3$, we have $$z_1 (z_2 z_3) = (z_1 z_2)z_3 = z_1 z_2 z_3.$$ L
- Question 9, Exercise 1.2
- \] For \(z_1 = 7 + 2i\) and \(z_2 = 3 - i\), we have: \[x_1 = 7, \quad y_1 = 2, \quad x_2 = 3, \quad... Given \(z_1 = 4 + 2i\) and \(z_2 = 2 + 5i\), we have: \[x_1 = 4, \quad y_1 = 2, \quad x_2 = 2, \quad... ] For \(z_1 = 5 - 4i\) and \(z_2 = 5 + 4i\), we have: \[x_1 = 5, \quad y_1 = -4, \quad x_2 = 5, \qua... Given \(z_1 = 3 - 7i\) and \(z_2 = 2 + 5i\), we have: \[x_1 = 3, \quad y_1 = -7, \quad x_2 = 2, \quad
- Question 7, Exercise 1.4
- eta_2}\right)=\arg(\theta_1)-\arg(\theta_2),$$ we have \begin{align*} &\arg \left(\dfrac{1-iz}{1-z}\righ... -1}\left(\frac{A+B}{1-AB}\right), \end{align*} we have \begin{align*} & \tan^{-1} \left( \frac{\frac{y}{... -1}\left(\frac{A+B}{1-AB}\right), \end{align*} we have \begin{align*} &\tan^{-1}\left(\dfrac{\frac{y-1}{
- Question 4, Exercise 1.1
- 2\\ -y&=-2\\ y&=2\end{align} Putting in $(3)$, we have $x=-2$. Hence $x=-2$ and $y=2$. GOOD ====Ques... -2x+1=0\cdots\cdots(2) \end{align} From $(2)$, we have &y=2x-1\cdots \cdots (3)\end{align} Put the value
- Question 5, Exercise 1.1
- .** Suppose $z=x+iy$, then $\bar{z}=x-iy$. So we have \begin{align}&4z-3\bar{z}=\dfrac{1-18i}{2-i}\\ \i... 7y &=-7 \,\text{ i.e. }\,y=-1.\end{align} Thus we have $z=x+iy=4-i$. GOOD ====Go to ==== <text align="
- Question 3, Exercise 1.2
- { or } \quad (\bar{z})^2=-y^2. ... (i)$$ Also, we have $$z^2=x^2 \quad \text{ or } \quad z^2=-y^2. ...(ii)$$ From (i) and (ii), we have $$(\overline{z})^{2}=z^{2}$$ Conversly, suppose t
- Question 8, Review Exercise
- $\theta=45^{\circ}$. ** Solution. ** Here we have $$x= \sqrt{2} + i \sqrt{2}, \quad \theta=\dfrac{\pi}{4}.$$ We have to find $x_{\max}$. By using the formula \begin{a
- Question 6, Exercise 1.2
- a+3=0. \end{align} By using quadratic formula, we have \begin{align} \lambda &=\dfrac{-3\pm\sqrt{9-4(1)
- Question 1, Exercise 1.4
- * Solution. ** Let $z=x+iy=2 + i 2 \sqrt{3}$. We have \begin{align} r & = \sqrt{x^2 + y^2} = \sqrt{2^2
- Question 5, Exercise 1.4
- (a^2+b^2+c^2-ab-bc-ca), \end{align} Using (3), we have $$a^3+b^3+c^3-3abc=0$$ $$\implies a^3+b^3+c^3=3ab
- Question 8, Exercise 1.4
- gle is: $\dfrac{\pi}{4}$ ** Solution. ** Here we have $$x_{\max}=0.004, \quad \theta=\dfrac{\pi}{4}.$$