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This is the second of our many quizzes on complex numbers. Click on this to access more such competitive quizzes on all topics of Complex numbers.

Complex Number Quiz 2

mathematics

 Express the following in the form of $a + bi: \left( i \right)\left( -2i \right){{\left( -\dfrac{1}{4}i \right)}^{3}}$.

 Perform the necessary mathematical operations and substitute ${{i}^{2}}=-1$ to simplify.

 $\begin{align} & \text{We have, }\left( i \right)\left( -2i \right){{\left( -\dfrac{1}{4}i \right)}^{3}}\\ &\Rightarrow 2\times \dfrac{1}{4\times 4\times 4}\times {{i}^{5}}\\ &\Rightarrow \dfrac{1}{32}{{\left( {{i}^{2}} \right)}^{2}}i=\dfrac{1}{32}i\end{align}$

 Find the multiplicative inverse of the complex number $8-3i$.

 If $z$ is a complex number then the multiplicative inverse of $z$ is given by: ${{z}^{-1}}=\dfrac{\overline{z}}{{{\left| z \right|}^{2}}}$.

 $\begin{align} & \text{Let }z=8-3i\text{ Then, }\overline{z}=8+3i\text{ and }{{\left| z \right|}^{2}}={{8}^{2}}+{{\left( -3 \right)}^{2}}=73\\ & \text{Therefore, the multiplicative inverse is given by: }\\ & \Rightarrow {{z}^{-1}}=\dfrac{\overline{z}}{{{\left| z \right|}^{2}}}=\dfrac{8}{73}+\dfrac{3}{73}i\end{align}$

 Express the given complex number in the form of $a+ib : {{\left( 2-2i \right)}^{4}}$.

 Let ${{z}_{1}}$ and ${{z}_{2}}$ are two complex numbers, then ${{\left( {{z}_{1}}-{{z}_{2}} \right)}^{2}}={{\left( {{z}_{1}} \right)}^{2}}+{{\left( {{z}_{2}} \right)}^{2}}-2{{z}_{1}}{{z}_{2}}$ and ${{i}^{2}}=-1$ 

 $\begin{align} & \text{We have, } {{\left( 2-2i \right)}^{4}}\Rightarrow {{\left[ {{\left( 2-2i \right)}^{2}} \right]}^{2}}\\ &\Rightarrow {{\left[ {{2}^{2}}+{{\left( 2i \right)}^{2}}-8i \right]}^{2}}\\ &\Rightarrow {{\left[ -8i \right]}^{2}}\Rightarrow -64\end{align}$

 Express the following expression in the form of $a+ib : \dfrac{\left( 2+i\sqrt{3} \right)\left( 2-i\sqrt{3} \right)}{\left( \sqrt{2}+i\sqrt{3} \right)-\left( \sqrt{2}-i\sqrt{3} \right)}$

 Let ${{z}_{1}}$ and ${{z}_{2}}$ are two complex numbers, then ${{\left( {{z}_{1}} \right)}^{2}}-{{\left( {{z}_{2}} \right)}^{2}}=\left( {{z}_{1}}+{{z}_{2}} \right)\left( {{z}_{1}}-{{z}_{2}} \right)$ and ${{i}^{2}}=-1$.

 $\begin{align} & \text{We have, }\dfrac{\left( 2+i\sqrt{3} \right)\left( 2-i\sqrt{3} \right)}{\left( \sqrt{2}+i\sqrt{3} \right)-\left( \sqrt{2}-i\sqrt{3} \right)}\\ &\Rightarrow \dfrac{{{2}^{2}}-{{\left( i\sqrt{3} \right)}^{2}}}{\sqrt{2}+i\sqrt{3}-\sqrt{2}+i\sqrt{3}}=\dfrac{7}{2\sqrt{3}i}\\ &\Rightarrow \dfrac{7}{2\sqrt{3}i}\times \dfrac{i}{i}=\dfrac{-7i}{2\sqrt{3}}\times \dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{-7\sqrt{3}i}{6}\end{align}$

 Given two complex numbers, ${{z}_{1}}=\left( 6-3i \right)$and ${{z}_{2}}=\left( 2+i \right)$ then find $\dfrac{{{z}_{1}}}{{{z}_{2}}}$.

 Let ${{z}_{1}}$ and ${{z}_{2}}$ are two complex numbers, where ${{z}_{2}}\ne 0$, the quotient $\dfrac{z_1}{z_2}={{z}_{1}}\dfrac{1}{{{z}_{2}}}$ and $z_{2}^{-1}=\dfrac{\overline{{{z}_{2}}}}{{{\left| {{z}_{2}} \right|}^{2}}}$, ${{i}^{2}}=-1$.

 $\begin{align} & \text{Given complex numbers: }{{z}_{1}}=\left( 6-3i \right)\text{ and }{{z}_{2}}=\left( 2+i \right)\\ &\Rightarrow \dfrac{{{z}_{1}}}{{{z}_{2}}}={{z}_{1}}\dfrac{1}{{{z}_{2}}}={{z}_{1}}.z_{2}^{-1}\\ &\Rightarrow z_{2}^{-1}=\dfrac{2-i}{{{2}^{2}}+{{1}^{2}}}=\dfrac{2}{5}-\dfrac{i}{5}\\ &\Rightarrow \dfrac{{{z}_{1}}}{{{z}_{2}}}=\left( 6-3i \right)\left( \dfrac{2}{5}-\dfrac{i}{5} \right)=\dfrac{1}{5}\left( 9-12i \right)\end{align}$