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- Question 4, Exercise 1.3
- {7}{53}i\end{align} Put value of $\omega$ in (1), we have \begin{align} &(1-i) z+(1+i)\left(\dfrac{2}{... frac{155}{106}+\dfrac{145}{106}i\end{align} Thus, we have $$z=\dfrac{155}{106}+\dfrac{145}{106}i, \ome... ts(2) \end{align} Multiplying $(1)$ by $(1-2i)$, we get: \begin{align} &(1-2i)(2i z) + (1-2i)(3-2i) ... 3) \end{align} Multiplying equation (2) by $2i$, we get: \begin{align} &2i(1-2i) z + 2i(3+2i) \omega
- Question 9, Exercise 1.2
- **Solution.** Suppose $z=3 - \sqrt{-4}=3-2i$. We will use the following formulas: \[\text{Re}(z^{... \dfrac{7+2 i}{3-i}\right)^{-1}$. **Solution.** We use the following formulas: \[Re\left(\left(\fr... 2}.\] For \(z_1 = 7 + 2i\) and \(z_2 = 3 - i\), we have: \[x_1 = 7, \quad y_1 = 2, \quad x_2 = 3, ... frac{4+2 i}{2+5 i}\right)^{-2}$. **Solution.** We will use the following formulas: \begin{align}
- Question 10, Exercise 1.2
- \, -- (4) \end{align} From (1), (2), (3) and (4), we have: $$\left| z_1 \right| = \left| -z_1 \right| ... 10} + \frac{7}{10}i. \,\, -- (i) \end{align} Now, we have \begin{align} \overline{z_1} = -3 - 2i, \qua... + \frac{7}{10}i.\,\, -- (ii)$$ From (i) and (ii), we have \[ \overline{\left( \frac{z_1}{z_2} \right)}... n \[ z_1 = -3 + 2i, \quad z_2 = 1 - 3i. \] First we calculate \begin{align} z_1 z_2 &= (-3 + 2i)(1
- Question 1, Review Exercise
- rational \\ <btn type="link" collapse="a1">See Answer</btn><collapse id="a1" collapsed="true">%%(c)%%:... (d) no \\ <btn type="link" collapse="a2">See Answer</btn><collapse id="a2" collapsed="true">(b): two... $(1,1)$ \\ <btn type="link" collapse="a3">See Answer</btn><collapse id="a3" collapsed="true">(a): $(0... \bar{z}$ \\ <btn type="link" collapse="a4">See Answer</btn><collapse id="a4" collapsed="true">(d): $\b
- Question 8, Exercise 1.2
- Solution.** Given: $$|2z-i|=4.$$ Put $z=x+i y$, we have \begin{align} & |2(x+iy)-i|=4 \\ \implies & ... n.** Given: $$|z-1|=|\bar{z}+i|.$$ Put $z=x+iy$, we have \begin{align} & |(x+iy)-1| = |(x-iy)+i| \\ \... qrt{x^2+(y-1)^2} \end{align} Squaring both sides, we get \begin{align} & (x-1)^2 + y^2 = x^2 + (y-1)^2... iven: $$|z-4i| + |z+4i| = 10.$$ Put $z = x + iy$, we have \begin{align} & |(x + iy) - 4i| + |(x + iy)
- Question 3, Exercise 1.4
- =\tan^{-1}\left(\dfrac{b}{a}\right). \end{align*} We can write these complex numbers in polar form as:... quad z=|z|e^{i\theta} \,\,-- (1) \end{align*} Now we have given \begin{align*} & \left(x_{1}+i y_{1}\r... z_3 \cdots z_n = z. \end{align*} By using $(1)$, we have \begin{align*} &|z_1| e^{i\theta_1}\cdot |z_... , -- (3) $$ Taking square on both sides of $(2)$, we get $$|z_1|^2\cdot|z_2|^2\cdot|z_3|^2\cdots |z_n|
- Question 2, Exercise 1.2
- ltiplicative assocative law} \end{align} That is, we have proved $$(z_1 z_2)(z_3 z_4)=(z_1 z_3) (z_2 z... tiplicative associative law} \end{align} That is, we have proved $$(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
- Question 5, Exercise 1.4
- \gamma). \\ \end{align} Using (1) and (2) above, we get $$a+b+c=0 -- (3).$$ Since we know \begin{align} &a^3+b^3+c^3-3abc \\ =&(a+b+c)(a^2+b^2+c^2-ab-bc-ca), \end{align} Using (3), we have $$a^3+b^3+c^3-3abc=0$$ $$\implies a^3+b^3+c^3=3abc.$$ Putting values of $a$, $b$ and $c$, we get \begin{align} &(e^{i\alpha})^3+(e^{i\beta})^3
- Question 3, Exercise 1.2
- \, ... (1)$$ First suppose that $z$ is real, then we shall prove $\overline{z}=z$. Since $z$ is real,... Now conversly suppose that $\overline{z}=z$, then we $z$ is real. As \begin{align}& z=\bar{z}\\ \Righ... ext{ or } \quad (\bar{z})^2=-y^2. ... (i)$$ 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, supp
- Question 7, Exercise 1.4
- \theta_2}\right)=\arg(\theta_1)-\arg(\theta_2),$$ we have \begin{align*} &\arg \left(\dfrac{1-iz}{1-z}... x}{1+y}\right)\\ \end{align*} Using it in $(1)$, we get \begin{align*} \tan^{-1}\left(\dfrac{y}{1-x}\... n^{-1}\left(\frac{A+B}{1-AB}\right), \end{align*} we have \begin{align*} & \tan^{-1} \left( \frac{\fra... n^{-1}\left(\frac{A+B}{1-AB}\right), \end{align*} we have \begin{align*} &\tan^{-1}\left(\dfrac{\frac{
- Question 6(i-ix), Exercise 1.4
- i \end{align*} <fc #ff0000>Note:</fc> Generally, we write exact answers, not approximate answers. =====Question 6(viii)===== Write a given complex number in the algebraic
- Question 4, Exercise 1.1
- &=-2\\ -y&=-2\\ y&=2\end{align} Putting in $(3)$, we have $x=-2$. Hence $x=-2$ and $y=2$. GOOD ===... &y-2x+1=0\cdots\cdots(2) \end{align} From $(2)$, we have &y=2x-1\cdots \cdots (3)\end{align} Put the
- Question 5, Exercise 1.1
- ion.** Suppose $z=x+iy$, then $\bar{z}=x-iy$. So we have \begin{align}&4z-3\bar{z}=\dfrac{1-18i}{2-i}... ad 7y &=-7 \,\text{ i.e. }\,y=-1.\end{align} Thus we have $z=x+iy=4-i$. GOOD ====Go to ==== <text al
- Question 4, Review Exercise
- ^2 + (y - 2)^2} \end{align*} Squaring both sides, we have \begin{align*} & x^2 + (y + 2)^2 = x^2 + (y ... \implies & 8y=0 \implies y=0. \end{align*} Hence, we conclude $z=x+i \cdot 0$ $\implies z=x$. ====G
- Question 8, Review Exercise
- hen $\theta=45^{\circ}$. ** Solution. ** Here we have $$x= \sqrt{2} + i \sqrt{2}, \quad \theta=\dfrac{\pi}{4}.$$ We have to find $x_{\max}$. By using the formula \be