Frac{{3 + 2isin θ }}{{1 - 2isin θ }} will be purely imaginary, if θ = [Where n is an integ

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4-1.Complex numbers

medium

$\frac{{3 + 2i\sin \theta }}{{1 - 2i\sin \theta }}$ will be purely imaginary, if $\theta = $

[Where $n$ is an integer]

$2n\pi \pm \frac{\pi }{3}$

$n\pi + \frac{\pi }{3}$

$n\pi \pm \frac{\pi }{3}$

None of these

(IIT-1976)

Solution

(c) $\frac{{3 + 2i\sin \theta }}{{1 – 2i\sin \theta }}$ will be purely imaginary, if $\theta = $will be purely imaginary, if the real part vanishes, i.e., $\frac{{3 – 4{{\sin }^2}\theta }}{{1 + 4{{\sin }^2}\theta }} = 0$
==> $3 – 4{\sin ^2}\theta = 0$ (only if $\theta $ be real)
==> $\sin \theta = \pm \frac{{\sqrt 3 }}{2} = \sin \left( { \pm \frac{\pi }{3}} \right)$
==> $\theta = n\pi + {( – 1)^n}\left( { \pm \frac{\pi }{3}} \right) = n\pi \pm \frac{\pi }{3}$

Standard 11

Mathematics

Let $S_{1}=\left\{z_{1} \in C:\left|z_{1}-3\right|=\frac{1}{2}\right\}$ and $S_{2}=\left\{z_{2} \in C:\left|z_{2}-\right| z_{2}+1||=\left|z_{2}+\right| z_{2}-1||\right\} . \quad$ Then, for $z_{1} \in S_{1}$ and $z_{2} \in S_{2}$, the least value of $\left|z_{2}-z_{1}\right|$ is.

hard

(JEE MAIN-2022)

If $|{a_k}| < 1,{\lambda _k} \ge 0$ for $k = 1,\,2,….n$ and ${\lambda _1} + {\lambda _2} + … + {\lambda _n} = 1,$ then the value of $|{\lambda _1}{a_1} + {\lambda _2}{a_2} + ….{\lambda _n}{a_n}|$ is

medium

If $a+i b=\frac{(x+i)^{2}}{2 x^{2}+1},$ prove that $a^{2}+b^{2}=\frac{\left(x^{2}+1\right)^{2}}{\left(2 x^{2}+1\right)^{2}}$

hard

Let complex number $z$ be such that $|z- \frac{6}{z}|=5$ then maximum value of $|z|$ will be –

normal

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

easy

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