The correct option is B 2i We have, ( 2i1+i)^2 ⇒4i^21+i^2+2i ⇒ 41 1+2i ( ∵i^2= 1 ) ⇒ 42i ⇒ 2i ⇒ 2ii^2 ⇒ 2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Properties of Iota

Solve: 2i1 ...

Question

Solve:
${(\frac{2 i}{1 + i})}^{2}$

- i

No worries! We‘ve got your back. Try BYJU‘S free classes today!

i

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 i

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1 - i

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $2 i$
We have,
${(\frac{2 i}{1 + i})}^{2}$
$\Rightarrow \frac{4 i^{2}}{1 + i^{2} + 2 i}$
$\Rightarrow \frac{- 4}{1 - 1 + 2 i} (∵ i^{2} = - 1)$
$\Rightarrow \frac{- 4}{2 i}$
$\Rightarrow \frac{- 2}{i}$
$\Rightarrow \frac{- 2 i}{i^{2}}$
$\Rightarrow \frac{- 2 i}{- 1}$
$\Rightarrow 2 i$

Hence, this is the answer.

Suggest Corrections

Similar questions

\infty \sum i = 1 \frac{i^{2}}{2^{i}}

{(\frac{1 + i}{1 - i})}^{2} + {(\frac{1 - i}{1 + i})}^{2}

z

(i - z) (1 + i) = 2 i

{(\frac{2 i}{1 + i})}^{2}

i

\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i; x, y \in R; i = \sqrt{- 1} .

Join BYJU'S Learning Program