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- Question 6, Exercise 1.3
- $ $$(z^2+z+1 )=0$$ By using quadratic formula, we have $$z=\dfrac{-1\pm \sqrt{1-4}}{2}$$ $$z=\dfrac{-1\p... $ $$(z^2-z+1 )=0$$ By using quadratic formula, we have $$z=\dfrac{1\pm \sqrt{1-4}}{2}$$ $$z=\dfrac{1\pm ... }i}{2}$$ The value of $z$ from both equations, we have $$z=\pm \dfrac{1}{2}\pm \dfrac{\sqrt{3}}{2}i$$ ... 2-2z+4=0$$ According to the quadratic formula, we have $a=1$, $b=-2$ and $c=4$ Thus, we have \begin{alig
- Question 5, Exercise 1.3
- }+z+3=0.$$ According to the quadratic formula, we have\\ $a=1$, $b=1$ and $c=3$.\\ Thus we have \begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ &=\... }}-z-1=0$$ According to the quadratic formula, we have\\ $a=1$, $b=-1$ and $c=-1$\\ So we have \begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ &=\d
- Question 5, Exercise 1.2
- &=3-i \ldots (2)\end{align} From (1) and (2), we have $$\overline{z_1+z_2}=\overline{z_1}+\overline{z_... \ &=13 \ldots (2)\end{align} From (1) and (2), we have $$\overline{z_1 z_2}=\overline{z_1}\overline{z_2}... {2}}} \ldots (2) \end{align} From (1) and (2), we have $$\overline{\left(\dfrac{z_1}{z_2}\right)}=\dfrac
- Question 11, Exercise 1.1
- dfrac{-2}{5}+\dfrac{11i}{5} \end{align} Hence, we have $${\rm Re}\left( \dfrac{{{z}_{1}}{{z}_{2}}}{\over... c{1}{z_1\overline{z_1}}=\dfrac{1}{5}.$$ Hence, we have $${\rm Im}\left(\dfrac{1}{z_1\overline{z_1}}\righ
- Question 2, Exercise 1.3
- z}^{3}}+6z+20$.\\ By using synthetic division, we have $$\begin{array}{c|cccc} -2 & 1 & 0 & 6 & 20 \\ ... {{z}^{2}}+z-2$.\\ By using synthetic division, we have $$\begin{array}{c|cccc} 2 & 1 & -2 & 1 & -2 \\
- Question 2, Exercise 1.2
- =4-3i \ldots (2) \end{align} From (1) and (2), we have the required result. Now, we prove associative p
- Question 6, Exercise 1.2
- native Method**\\ We know $|z|^2=z\bar{z}$, so we have \begin{align}|{{z}_{1}}{{z}_{2}}{{|}^{2}}&={{z}_{
- Question 8, Exercise 1.2
- $z=a+bi$ ... (1) Then $\overline{z}=a-bi$. We have given \begin{align}&z=\overline{z} \\ \implies &a
- Question 3 & 4, Exercise 1.3
- 5=0 \ldots (i)$$ Put $z=1+2i$ in equaiton (i), we have \begin{align}L.H.S.&=(1+2i)^2-2(1+2i)+5\\ &=1+(2
- Question 4 & 5, Review Exercise 1
- Solution==== Given $z_1=2-i$ and $z_2=1+i$, so we have \begin{align} \dfrac{{{z}_{1}}+{{z}_{2}}+1}{{{z}_
- Question 6, 7 & 8, Review Exercise 1
- end{align} According to the quadratic formula, we have\\ $$a=1,\quad b=-2$$ and $$c=2$$\\ Quadratic for