{left( frac{1 + i}{1 - i} right)^2} + {left( | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) First,simplify the term $\frac{1 + i}{1 - i}$ by multiplying the numerator and denominator by the conjugate $(1 + i)$:$\frac{1 + i}{1 - i} = \frac

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(C) First,simplify the term $\frac{1 + i}{1 - i}$ by multiplying the numerator and denominator by the conjugate $(1 + i)$:$\frac{1 + i}{1 - i} = \frac{(1 + i)(1 + i)}{(1 - i)(1 + i)} = \frac{1 + 2i + i^2}{1 - i^2} = \frac{1 + 2i - 1}{1 - (-1)} = \frac{2i}{2} = i$.Similarly,$\frac{1 - i}{1 + i} = \frac{1}{i} = -i$.Now,substitute these values into the expression:$(i)^2 + (-i)^2 = -1 + (-1)^2 = -1 + 1 = 0$.Wait,re-evaluating the expression:${\left( \frac{1 + i}{1 - i} \right)^2} + {\left( \frac{1 - i}{1 + i} \right)^2} = (i)^2 + (-i)^2 = -1 + (-1) = -2$.

${\left( \frac{1 + i}{1 - i} \right)^2} + {\left( \frac{1 - i}{1 + i} \right)^2}$ is equal to

Similar Questions

The imaginary part of $\frac{(1 + i)^2}{2 - i}$ is

If $\left|\begin{array}{cc}1-i & i \\ 1+2 i & -i\end{array}\right|=x+i y$, then $x$ is equal to

The value of $\{i^{22}-(\frac{1}{i})^{35}\}^2$ is

The expression $\frac{3 + 2i\sin \theta}{1 - 2i\sin \theta}$ will be real,if $\theta = $ [Where $n$ is an integer]

If $\sum\limits_{k = 0}^{100} {{i^k}} = x + iy$,then the values of $x$ and $y$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module