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

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

Question

(B) First,simplify the fraction inside the bracket by multiplying the numerator and denominator by the conjugate of the denominator $(1-i)$:$\frac{2i}

Vedclass · Accepted Answer

$2i$

Vedclass · Answer

(B) First,simplify the fraction inside the bracket by multiplying the numerator and denominator by the conjugate of the denominator $(1-i)$:$\frac{2i}{1+i} = \frac{2i(1-i)}{(1+i)(1-i)} = \frac{2i - 2i^2}{1 - i^2} = \frac{2i + 2}{1 + 1} = \frac{2(i+1)}{2} = i+1$.Now,square the result:$(i+1)^2 = i^2 + 1^2 + 2i = -1 + 1 + 2i = 2i$.Therefore,the correct option is $B$.

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

Similar Questions

Express the following in the form of $a+bi$: $(-i)(2i)\left(-\frac{1}{8}i\right)^{3}$

If $i = \sqrt{-1}$, then $[i^{18} + (\frac{1}{i})^{25}]^3 = $

The statement $(a + ib) < (c + id)$ is true for

If $n$ is a positive integer,then $\left( \frac{1 + i}{1 - i} \right)^{4n + 1} = $

$\frac{(1+i)^{2011}}{(1-i)^{2009}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module