Question

The value of $(1+\mathrm{i}{)}^5(1-\mathrm{i}{)}^5$

• A. $-8$
• B. $8\mathrm{i}$
• C. $8$
• D. $32$

Answer 1

The correct option is D

$32$

Explanation for the correct option.

Find the value

Given: $(1+\mathrm{i}{)}^5(1-\mathrm{i}{)}^5$ $={\left(1+\mathrm{i}\right)}^4\left(1+\mathrm{i}\right){\left(1-\mathrm{i}\right)}^4\left(1-\mathrm{i}\right)-\left(1\right)$

$$\begin{gathered} {\left(1+\mathrm{i}\right)}^2=1^2+2\times 1\times \mathrm{i}+{\mathrm{i}}^2=1+2\mathrm{i}-1=2\mathrm{i}\left[{(\mathrm{a}+\mathrm{b})}^2={\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab}\right] \\ {\left(1-\mathrm{i}\right)}^2=1^2-2\times 1\times \mathrm{i}+{\mathrm{i}}^2=1-2\mathrm{i}-1=-2\mathrm{i}\left[{(\mathrm{a}+\mathrm{b})}^2={\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}\right] \end{gathered}$$

Put the above value in (1)

$$\begin{gathered} \Rightarrow {\left(2\mathrm{i}\right)}^2\left(1+\mathrm{i}\right){\left(-2\mathrm{i}\right)}^2\left(1-\mathrm{i}\right) \\ =4{\mathrm{i}}^2\times 4{\mathrm{i}}^2\times \left(1-{\mathrm{i}}^2\right)\left[(\mathrm{a}-\mathrm{b})(\mathrm{a}+\mathrm{b})=({\mathrm{a}}^2-{\mathrm{b}}^2)\right] \\ =-4\times -4\times \left(1+1\right)\left[{\mathrm{i}}^2=-1\right] \\ =16\times 2 \\ =32 \end{gathered}$$

Hence $o\mathrm{ption}\left(\mathrm{D}\right)$ is the correct answer.