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- Question 8, Exercise 1.2
- &={{a}^{2}}-{{bi}^{2}}\\ &={{a}^{2}}-b^2 (-1)\\ \implies z\overline{z}&={{a}^{2}}+b^2. \ldots (1) \end{al... . We have given \begin{align}&z=\overline{z} \\ \implies &a+bi=a-bi \\ \implies &bi=bi \\ \implies &2bi=0 \\ \implies &b=0 \end{align} Using it in (1), we get $z=a+0i=a$, that is, $z$ is
- Question 1, Exercise 1.3
- 1\\ \end{array} \] \begin{align}-11w&=11i-11\\ \implies w&=\dfrac{11-11i}{11}\\ \implies w&=1-i\end{align} Put value of $w$ in (i). \begin{align}z&-4(1-i)=3i\\ \implies z &=4(1-i)+3i\\ &=4-4i+3i\\ &=4-i\end{align} He... lue of $w$ in (i).\\ \begin{align}z&+(2-6i)=3i\\ \implies z&=3i-2+6i\\ &=-2+9i\end{align} Hence $$z=-2+9i
- Question 6, Exercise 1.3
- z^3=-8.$$ This gives \begin{align} & z^3+2^3=0\\ \implies &(z+2)\left(z^2-2z+4 \right)=0 \\ & \quad \becaus... \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)\\ \implies & z+2=0 \quad \text{or} \quad z^2-2z+4=0.\end{ali... efore, we have \begin{align} &z^3-1-3z(z-1)=-1\\ \implies &{{z}^{3}}-1-3{{z}^{2}}+3z+1=0\\ \implies &{{z}^{3}}-3{{z}^{2}}+3z=0\\ \implies &z\left(z^2-3z+3 \right)=0
- Question 5, Exercise 1.3
- .\\ ====Solution==== Given: $${{z}^{2}}-1=z$$ $$\implies {{z}^{2}}-z-1=0$$ According to the quadratic form... tion==== Given: $${{z}^{2}}+4=0$$ \begin{align} \implies &z^2-(2i)^2=0\\ \implies &(z+2i)(z-2i)=0\\ \implies &z+2i=0 \quad \text{or} \quad z-2i=0\\ \implies &z=-2i \quad \text{or} \quad z=
- Question 3 & 4, Exercise 1.3
- +2\\ &=1-2i-1-2+2i+2\\ &=0=R.H.S\end{align} This implies $z_1=-1+i$ satisfied the given equation.\\ Now pu... )+2\\ &=1+2i-1-2-2i+2\\ &=0=R.H.S\end{align} This implies $z_2=-1-i$ satisfied the equation. =====Question... +(2i)^2+4i-2-4i+5\\ &=1-4+5\\ &=0\end{align} This implies $1+2i$ is solution of the given equation. ====Go
- Question 9, Exercise 1.2
- lign} &-\sqrt{13} \leq \sqrt{9} \leq \sqrt{13}\\ \implies &-|z|\leq \operatorname{Re}\left( z \right)\leq |... lign} &-\sqrt{13} \leq \sqrt{4} \leq \sqrt{13}\\ \implies &-|z|\leq {\rm Im}(z) \leq |z| \end{align} ==== G
- Question 6, Exercise 1.2
- {1}{\sqrt{{{a}^{2}}+{{b}^{2}}}}=\dfrac{1}{|z|}\\ \implies \left|\dfrac{1}{z} \right|&=\dfrac{1}{|z|} … (1)\
- Question 2, Exercise 1.3
- \\ \hline & 1 & 0 & 1 & 0 \\ \end{array}$$ This implies \begin{align}P(z)&=(z-2)\left( {{z}^{2}}+1 \right