Smallest positive integer n for which {(1 + i)^{2n}} = {(1

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4-1.Complex numbers

normal

The smallest positive integer $n$for which ${(1 + i)^{2n}} = {(1 - i)^{2n}}$is

$1$

$2$

$3$

$4$

Solution

(b) We have ${(1 + i)^{2n}} = {(1 – i)^{2n}}$
$ \Rightarrow {\left( {\frac{{1 + i}}{{1 – i}}} \right)^{2n}} = 1$$ \Rightarrow {(i)^{2n}} = 1$$ \Rightarrow {(i)^{2n}} = {( – 1)^2}$
$ \Rightarrow {(i)^{2n}} = {({i^2})^2}$$ \Rightarrow {(i)^{2n}} = {(i)^4}$

$ \Rightarrow 2n = 4$$ \Rightarrow n = 2$.

Standard 11

Mathematics

Let $z_{1}=2-i, z_{2}=-2+i .$ Find

$ \operatorname{Im}\left(\frac{1}{z_{1} \bar{z}_{1}}\right)$

medium

If $\frac{{{{(p + i)}^2}}}{{2p – i}} = \mu + i\lambda ,$then ${\mu ^2} + {\lambda ^2}$ is equal to

medium

The true statement is

normal

Let $z\,\ne -i$ be any complex number such that $\frac{{z – i}}{{z + i}}$ is a purely imaginary number. Then $z +\frac {1}{z}$ is

hard

(JEE MAIN-2014)

If $z \neq 0$ be a complex number such that $\left| z -\frac{1}{ z }\right|=2$, then the maximum value of $|z|$ is.

hard

(JEE MAIN-2022)