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  4. If z=r(cos θ+i sin θ), then the value of (z/ barz)+( barz/z) is

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Q. If $z=r(cos\, \theta+i \,sin \,\theta)$, then the value of $\frac{z}{\bar{z}}+\frac{\bar{z}}{z}$ is

Complex Numbers and Quadratic Equations

Solution:

Given, $z=r(cos \,\theta+i \,sin \,\theta) ;$
$ \bar{z}=r(\cos \theta-i \sin \theta)$
$\frac{z}{\bar{z}}=\frac{r(\cos \theta+i \sin \theta)}{r(\cos \theta-i \sin \theta)}$
$=(cos \,\theta+i\,sin\, \theta)(cos \,\theta-i\, \sin \theta)^{-1}$
$=(\cos \theta+i \sin \theta)(\cos \theta+i \sin \theta)$
$=(\cos \theta+i \sin \theta)^{2}=\cos 2 \theta+i \sin 2 \theta$
$ \therefore \frac{\bar{z}}{z}=(cos \,\theta-i \,\sin\, \theta)(\cos \theta+i \sin \theta)^{-1} $
$=(\cos \theta-i \sin \theta)(\cos \theta-i \sin \theta)$
$=(\cos \theta-i \sin \theta)^{2}=(\cos 2 \theta-i \sin 2 \theta) $
$ \therefore \frac{z}{\bar{z}}+\frac{\bar{z}}{z}=\cos 2 \theta+i \sin 2 \theta+\cos 2 \theta-i \sin 2 \theta$
$=2 \, cos \,2 \theta $