Question

The conjugate of the complex number $\frac{{\left(1+i\right)}^2}{\left(1-i\right)}$ is

• A. $1-i$
• B. $1+i$
• C. $-1+i$
• D. $-1-i$

Answer 1

The correct option is D

$-1-i$

Explanation for the correct option:

Calculating the conjugate of the given complex number:

Given complex number is $\frac{{\left(1+i\right)}^2}{\left(1-i\right)}$

First rationalize the given complex number;

Multiply $\left(1+i\right)$ in the numerator and denominator.

$\frac{{\left(1+i\right)}^2}{\left(1-i\right)}=\frac{{\left(1+i\right)}^2}{\left(1-i\right)}\times \frac{(1+i)}{(1+i)}$

Expanding the RHS we get,

$$\begin{gathered} =\frac{(1+2i-1)(1+i)}{1^2-i^2} \\ =\frac{2i(1+i)}{1-{(-1)}^2} \\ =\frac{2i-2i^2}{2} \\ =\frac{2i-2}{2} \\ =i-1 \end{gathered}$$

Therefore, the conjugate of the complex number is $(-1-i)$

Hence, option (D) is the correct answer.