Answer :

$ \begin{matrix} (5 i)(\dfrac{-3}{5} i) & =-5 \times \dfrac{3}{5} \times i \times i & \\ & =-3 i^{2} \\ & =-3(-1) & {[i^{2}=-1]} \\ & =3 & \end{matrix} $

Answer :

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

Answer :

$ \begin{matrix} i^{-39} & =i^{-4 \times 9-3}=(i^{4})^{-9} \cdot i^{-3} \\ & =(1)^{-9} \cdot i^{-3} & {[i^{4}=1]} \\ & =\dfrac{1}{i^{3}}=\dfrac{1}{-i} & {[i^{3}=-i]} \\ & =\dfrac{-1}{i} \times \dfrac{i}{i} & \\ & =\dfrac{-i}{i^{2}}=\dfrac{-i}{-1}=i & & {[i^{2}=-1]} \end{matrix} $

Answer :

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

Answer :

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

Answer :

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

Answer :

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

Answer :

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

$=\dfrac{1}{3}+\dfrac{7}{3} i+4+\dfrac{1}{3} i+\dfrac{4}{3}-i$

$=(\dfrac{1}{3}+4+\dfrac{4}{3})+i(\dfrac{7}{3}+\dfrac{1}{3}-1)$

$=\dfrac{17}{3}+i \dfrac{5}{3}$

Answer :

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

Answer :

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

Find the multiplicative inverse of each of the complex numbers given in the Exercises 11 to 13.

Answer :

Let $z=4 -{3 i}$

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

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 $|z|^{2}=(\sqrt{5})^{2}+3^{2}=5+9=14$

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{(3+i \sqrt{5})(3-i \sqrt{5})}{(\sqrt{3}+\sqrt{2} i)-(\sqrt{3}-i \sqrt{2})} $

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

$=\dfrac{9-5 i^{2}}{2 \sqrt{2} i}$

$=\dfrac{9-5(-1)}{2 \sqrt{2} i} \quad[i^{2}=-1]$

$=\dfrac{9+5}{2 \sqrt{2} i} \times \dfrac{i}{i}$

$=\dfrac{14 i}{2 \sqrt{2} i^{2}}$

$=\dfrac{14 i}{2 \sqrt{2}(-1)}$

$=\dfrac{-7 i}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}}$

$=\dfrac{-7 \sqrt{2} i}{2}$