Question

Find the real part of ${\tan }^{-1}\left(1+i\right)$.

• A. $-\frac{{\tan }^{-1}\left(2\right)}{2}$
• B. $\frac{{\tan }^{-1}\left(2\right)}{2}$
• C. $-\frac{{\tan }^{-1}\left({\displaystyle \frac{1}{2}}\right)}{2}$
• D. $0$

Answer 1

The correct option is A

$-\frac{{\tan }^{-1}\left(2\right)}{2}$

Simplifying the given function using Trigonometric identities

Let,

$$\begin{gathered} {\tan }^{-1}\left(1+i\right)=a+ib \\ {\tan }^{-1}\left(1-i\right)=a-ib \end{gathered}$$

Therefore,

$$\begin{gathered} {\tan }^{-1}\left(1+i\right)+{\tan }^{-1}\left(1-i\right)=a+ib+a-ib \\ \Rightarrow {\tan }^{-1}\left(\frac{1+i+1-i}{1-(1+i)(1-i)}\right)=2a \\ \Rightarrow {\tan }^{-1}\left(\frac{2}{1-1+i^2}\right)=2a \\ \Rightarrow {\tan }^{-1}\left(-2\right)=2a \\ \Rightarrow a=-\frac{{\tan }^{-1}\left(2\right)}{2} \end{gathered}$$

The real part of ${\tan }^{-1}\left(1+i\right)$ is $a$.

Therefore, Option(A) is the correct answer.