Answer :

$ \begin{aligned} \left(5 i\right)\left(\dfrac{-3}{5} i\right) & =-5 \times \dfrac{3}{5} \times i \times i & \\ \\ & =-3 i^{2} \\ \\ & =-3\left(-1\right) \qquad {\left[\because \ \ i^{2}=-1\right]} \\ \\ & =3 & \end{aligned} $

Answer :

$\large{ \begin{aligned} i^{9}+i^{19} & =i^{4 \times 2+1}+i^{4 \times 4+3} \\ \\ & =\left(i^{4}\right)^{2} \cdot i+\left(i^{4}\right)^{4} \cdot i^{3} \\ \\ & =1 \times i+1 \times\left(-i\right) \quad\left[\because \ \ i^{4}=1, i^{3}=-i\right] \\ \\ & =i+\left(-i\right) \\ \\ & =0 \end{aligned}} $

Answer :

$\large{ \begin{aligned} i^{-39} & =i^{-4 \times 9-3}=\left(i^{4}\right)^{-9} \cdot i^{-3} \\ \\ & =\left(1\right)^{-9} \cdot i^{-3} \qquad {\left[\because \ \ i^{4}=1\right]} \\ \\ & =\dfrac{1}{i^{3}}=\dfrac{1}{-i} \qquad {\left[\because \ \ i^{3}=-i\right]} \\ \\ & =\dfrac{-1}{i} \times \dfrac{i}{i} & \\ \\ & =\dfrac{-i}{i^{2}}=\dfrac{-i}{-1}=i \qquad {\left[\because \ \ i^{2}=-1\right]} \end{aligned}} $

Answer :

$\large{ \begin{aligned} 3\left(7+i 7\right)+i\left(7+i 7\right) & =21+21 i+7 i+7 i^{2} & \\ \\ & =21+28 i+7 \times\left(-1\right) & \\ \\ & =14+28 i & \end{aligned}} $

Answer :

$\large{ \begin{aligned} \left(1-i\right)-\left(-1+i 6\right) & =1-i+1-6 i \\ \\ & =2-7 i \end{aligned}} $

Answer :

$\large{ \begin{aligned} \left(\dfrac{1}{5}+i \dfrac{2}{5}\right)-\left(4+i \dfrac{5}{2}\right) & =\dfrac{1}{5}+\dfrac{2}{5} i-4-\dfrac{5}{2} i \\ \\ & =\left(\dfrac{1}{5}-4\right)+i\left(\dfrac{2}{5}-\dfrac{5}{2}\right) \\ \\ & =\dfrac{-19}{5}+i\left(\dfrac{-21}{10}\right) \\ \\ & =\dfrac{-19}{5}-\dfrac{21}{10} i \end{aligned}} $

Answer :

$\left[\left(\dfrac{1}{3}+i \dfrac{7}{3}\right)+\left(4+i \dfrac{1}{3}\right)\right]-\left(\dfrac{-4}{3}+i\right)=\dfrac{1}{3}+\dfrac{7}{3} i+4+\dfrac{1}{3} i+\dfrac{4}{3}-i$

$\hspace{6.1cm}=\left(\dfrac{1}{3}+4+\dfrac{4}{3}\right)+i\left(\dfrac{7}{3}+\dfrac{1}{3}-1\right)$

$\hspace{6.1cm}=\dfrac{17}{3}+i \dfrac{5}{3}$

Answer :

$\large{ \begin{aligned} \left(1-i\right)^{4} & =\left[ \left(1-i\right)^{2}\right]^{2} \\ \\ & =\left[ 1^{2}+i^{2}-2 i\right]^{2} \\ \\ & =\left[ 1-1-2 i\right]^{2} \\ \\ & =\left(-2 i\right)^{2} \\ \\ & =\left(-2 i\right) \times\left(-2 i\right) \\ \\ & =4 i^{2}=-4 \quad\left[\because \ \ i^{2}=-1\right] \end{aligned}} $

Answer :

$ \begin{aligned} \left(\dfrac{1}{3}+3 i\right)^{3} & =\left(\dfrac{1}{3}\right)^{3}+\left(3 i\right)^{3}+3\left(\dfrac{1}{3}\right)\left(3 i\right)\left(\dfrac{1}{3}+3 i\right) \\ \\ & =\dfrac{1}{27}+27 i^{3}+3 i\left(\dfrac{1}{3}+3 i\right) & \\ \\ & =\dfrac{1}{27}+27\left(-i\right)+i+9 i^{2} \qquad {\left[\because \ \ i^{3}=-i\right]} \\ \\ & =\dfrac{1}{27}-27 i+i-9 & \\ \\ & =\left(\dfrac{1}{27}-9\right)+i\left(-27+1\right) & \\ \\ & =\dfrac{-242}{27}-26 i & \end{aligned} $

Answer :

$ \begin{aligned} \left(-2-\dfrac{1}{3} i\right)^{3} & =\left(-1\right)^{3}\left(2+\dfrac{1}{3} i\right)^{3} & \\ \\ & =-\left[ 2^{3}+\left(\dfrac{i}{3}\right)^{3}+3\left(2\right)\left(\dfrac{i}{3}\right)\left(2+\dfrac{i}{3}\right)\right] \\ \\ & =-\left[8+\dfrac{i^{3}}{27}+2 i\left(2+\dfrac{i}{3}\right)\right] \qquad {\left[\because \ \ i^{3}=-i\right]} \\ \\ & =-\left[ 8-\dfrac{i}{27}+4 i+\dfrac{2 i^{2}}{3}\right] \qquad {\left[\because \ \ i^{2}=-1\right]} \\ \\ & =-\left[ 8-\dfrac{i}{27}+4 i-\dfrac{2}{3}\right] & \\ \\ & =-\left[ \dfrac{22}{3}+\dfrac{107 i}{27}\right] & \end{aligned} $

Answer :

Let $z=4 -{3 i}$

Then, $\bar{{}z}=4+3 i$ and $|z|^{2}=4^{2}+\left(-3\right)^{2}=16+9=25$

Therefore, the multiplicative inverse of $4- 3 i$ is given by

$ z^{-1}=\dfrac{\bar{{}z}}{|z|^{2}}=\dfrac{4+3 i}{25}=\dfrac{4}{25}+\dfrac{3}{25} i $

Answer :

Let $z=\sqrt{5}+3 i$

Then, $\bar{{}z}=\sqrt{5}-3 i$ and $|z|^{2}=\left(\sqrt{5}\right)^{2}+3^{2}=5+9=14$

Therefore, the multiplicative inverse of $\sqrt{5}+3 i$ is given by

$z^{-1}=\dfrac{\bar{{}z}}{|z|^{2}}=\dfrac{\sqrt{5}-3 i}{14}=\dfrac{\sqrt{5}}{14}-\dfrac{3 i}{14}$

Answer :

Let $z=- i$

Then, $\bar{{}z}=i$ and $|z|^{2}=1^{2}=1$

Therefore, the multiplicative inverse of $- i$ is given by

$z^{-1}=\dfrac{\bar{{}z}}{|z|^{2}}=\dfrac{i}{1}=i$

Answer :

$ \dfrac{\left(3+i \sqrt{5}\right)\left(3-i \sqrt{5}\right)}{\left(\sqrt{3}+\sqrt{2} i\right)-\left(\sqrt{3}-i \sqrt{2}\right)} =\dfrac{\left(3\right)^{2}-\left(i \sqrt{5}\right)^{2}}{\sqrt{3}+\sqrt{2} i-\sqrt{3}+\sqrt{2} i} \quad\left[\because \ \ \left(a+b\right)\left(a-b\right)=a^{2}-b^{2}\right]$

$\hspace{4.2cm}=\dfrac{9-5 i^{2}}{2 \sqrt{2} i}=\dfrac{9-5\left(-1\right)}{2 \sqrt{2} i} \quad\left[\because \ \ i^{2}=-1\right]$

$\hspace{4.2cm}=\dfrac{9+5}{2 \sqrt{2} i} \times \dfrac{i}{i}=\dfrac{14 i}{2 \sqrt{2} i^{2}}$

$\hspace{4.2cm}=\dfrac{14 i}{2 \sqrt{2}\left(-1\right)}=\dfrac{-7 i}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}}$

$\hspace{4.2cm}=\dfrac{-7 \sqrt{2} i}{2}$