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Question 5, Exercise 1.2
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z}_{1}}}+\overline{{{z}_{2}}}$. ====Solution==== Given ${{z}_{1}}=2+4i$ and ${{z}_{2}}=1-3i$. Thus $\ov... {z}_{1}}}\overline{{{z}_{2}}}$. ====Solution==== Given ${{z}_{1}}=2+3i$ and ${{z}_{2}}=2-3i$\\ Thus $\ov... _{1}}}}{\overline{{{z}_{2}}}}$. ====Solution==== Given ${{z}_{1}}=-a-3bi$ and ${{z}_{2}}=2a-3bi$.\\ Thus
Question 1, Exercise 1.3
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=3i\\ &2z+3w=11-5i\end{align} ====Solution==== Given that \begin{align}z-4w&=3i …(i)\\ 2z+3w&=11-5i ... n}&z+w=3i\\ &2z+3w=2\end{align} ====Solution==== Given that \begin{align}z+w&=3i …(i)\\ 2z+3w&=2 …(ii... -i\\ &(2-i)z-w=-1+i\end{align} ====Solution==== Given that \begin{align}3z+\left( 2+i \right)w&=11-i
Question 11, Exercise 1.1
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{\overline{{{z}_{1}}}} \right)$. ====Solution==== Given $z_1=2-i$ and $z_2=-2+i$, then $\overline{z_1}=2+... \overline{{{z}_{1}}}} \right)$. ====Solution==== Given $z_1=2-i$, then $\overline{z_1}=2+i$. \begin{alig
Question 3 & 4, Exercise 1.2
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e of the complex number $5+2i$. ====Solution==== Given $z=5+2i$. Here $a=5$ and $b=2$. Additive inverse... ex number $\left( 7,-9 \right)$. ====Solution==== Given $z=(7,-9)=7-9i$. Here $a=7$ and $b=-9$. Additive
Question 9, Exercise 1.2
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e{Re}\left( z \right)\leq |z|$ ====Solution==== Given $z=3+2i$. Then $|z|=\sqrt{9+4}=\sqrt{13}$ and ${\... e{Im}\left( z \right)\leq |z|$ ====Solution==== Given $z=3+2i$. Then $|z|=\sqrt{9+4}=\sqrt{13}$ and ${\
Question 2, Exercise 1.3
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ft( z \right)={{z}^{3}}+6z+20$$ ====Solution==== Given: $$p\left( z \right)={{z}^{3}}+6z+20$$ By factor... (z)={{z}^{3}}-2{{z}^{2}}+z-2.$$ ====Solution==== Given: $$P\left( z \right)={{z}^{3}}-2{{z}^{2}}+z-2$$ \
Question 3 & 4, Exercise 1.3
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equation ${{z}^{2}}+2z+2=0$\\ ====Solution==== Given: $$z^2+2z_1+2=0\quad \ldots (i)$$ Put the value o... end{align} This implies $z_1=-1+i$ satisfied the given equation.\\ Now put $z_2=-1-i$ in (i) \begin{alig
Question 1, Exercise 1.2
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.t. addition and multiplication. ====Solution==== Given ${{z}_{1}}=2+i$, ${{z}_{2}}=1-i$. First, we prove
Question 2, Exercise 1.2
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t. addition and multiplication. ====Solution==== Given ${{z}_{1}}=-1+i$, ${{z}_{2}}=3-2i$ and ${{z}_{3}}
Question 8, Exercise 1.2
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+bi$ ... (1) Then $\overline{z}=a-bi$. We have given \begin{align}&z=\overline{z} \\ \implies &a+bi=a-