1

Let $$z=\frac{3+2i\cos\theta}{1-3i\cos\theta},$$ such that $z$ is purely imaginary and $\theta \in\left(0, \frac{\pi}2\right) $, then find $$\sin^2(3\theta)+\cos^2{\theta}.$$

Let $z=ni$, where $n$ is any real number so

$ni=\frac{3+2i\cos\theta}{1-3i\cos\theta}$

then $ni+3n\cos\theta=3+2i\cos\theta$

on comparing both sides we get $n\cos\theta=1 \text{and } n= 2\cos\theta$

we get $\theta =\frac{π}{4}$ and thus the answer is $1.$

However, I was wondering if there’s an alternate way of solving this problem.

Anne Bauval
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math and physics forever
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2 Answers2

3

You can multiply by $1+3i\cos(\theta)$ eliminating the need for a 2nd variable

You get $\displaystyle z =\frac{3 - 6\cos^2(\theta) + 11i\cos(\theta)}{1 + 9\cos^2(\theta)} $

Since Re(z) = $0$

You directly get

$\displaystyle \cos(\theta)$ = $\displaystyle \frac{\pi}{4}$ + $\displaystyle n\frac{\pi}{2}$ , $\displaystyle n\in{R}$

No need to involve the 2nd variable

Jinansh Dalal
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1

$z$ is purely imaginary iff $(3+2i\cos\theta)(1+3i\cos\theta)$ is, i.e. iff $3-6\cos^2\theta=0,$ i.e. $$\cos^2\theta=\frac12,$$ i.e. $$\theta\in\frac\pi4+\Bbb Z\frac\pi2.$$ For every such $\theta$ (not necessarily $\in\left(0, \frac{\pi}2\right) $), $\sin^2(3\theta)+\cos^2{\theta}=1.$

Anne Bauval
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