Download App

STD 11 - 4.1 complex nubers — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 4.1 complex nubers1 Mark
MCQ
${\left\{ {\frac{{2i}}{{1 + i}}} \right\}^2} = $
  • A
    $1$
  • ✓
    $2i$
  • C
    $i-i$
  • D
    $1-2i $
✓

Answer

Correct option: B.
$2i$
b
(b)${\left[ {\frac{{2i}}{{1 + i}}} \right]^2} = {\left[ {\left( {\frac{{2i}}{{1 + i}}} \right)\,\left( {\frac{{1 - i}}{{1 - i}}} \right)} \right]^2} = {(i + 1)^2} = {i^2} + 1 + 2i = 2i$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from STD 11 - 4.1 complex nubers→STD 11 - 4.1 complex nubers — all question types→Maths STD 11 Science question papers→CBSE Board STD 11 Science question papers→CBSE Board question paper generator→

Similar questions

$3 \times 7^{22}+2 \times 10^{22}-44$ when divided by $18$ leaves the remainder .... .
→
Two vertices of a triangle are $(4, -3)$ and $(-2, 5)$. If the orthocentre of the triangle is at $(1, 2)$, then the third vertex is
→
If $\frac{1}{x} + x = 2\cos \theta ,$ then ${x^n} + \frac{1}{{{x^n}}}$ is equal to
→
$\frac{{{{(\cos \theta + i\sin \theta )}^4}}}{{{{(\sin \theta + i\cos \theta )}^5}}}$ =
→
$A(3, 2, 0), B(5, 3, 2), C(-9, 6, -3)$ are three points forming a triangle. If $AD,$ the bisector of $\angle\text{BAC}$ meets $BC$ in $D$ then coordinates of $D$ are:
→
If $(2\cos x - 1)(3 + 2\cos x) = 0,\,0 \le x \le 2\pi $, then $x = $
→
There are $12$ volleyball players in all in a college, out of which a team of $9$ players is to be formed. If the captain always remains the same, then in how many ways can the team be formed
→
If the roots of the equation $l{x^2} + nx + n = 0$ be in the ratio $p:q$ then $\sqrt {\frac{p}{q}} + \sqrt {\frac{q}{p}} + \sqrt {\frac{n}{l}} = $
→
‘$A$’ draws two cards with replacement from a pack of $52$ cards and ‘$B$' throws a pair of dice what is the chance that ‘$A$’ gets both cards of same suit and ‘$B$’ gets total of $6$
→
An urn contains marbles of four colours : red, white, blue and green. When four marbles are drawn without replacement, the following events are equally likely

$1.$ the selection of four red marbles.

$2.$ the selection of one white and three red marbles.

$3.$ the selection of one white, one blue and two red marbles.

$4.$ the selection of one marble of each colour.

The smallest total number of marbles satisfying the given condition is

→