# Find the two square roots of 36i.

Find the two square roots of 36i.
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getrdone07tl
$z=36i\phantom{\rule{0ex}{0ex}}=0+36i$
Now,
$r=\sqrt{{0}^{2}+\left(36{\right)}^{2}}\phantom{\rule{0ex}{0ex}}=36$
&
$\theta ={\mathrm{tan}}^{-1}\left(\frac{x}{y}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left(\frac{36}{0}\right)\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}\left(\mathrm{\infty }\right)\phantom{\rule{0ex}{0ex}}=\frac{\pi }{2}$
Now the polar form is,
$z=r\left(\mathrm{cos}\theta +i\mathrm{sin}\theta \right)\phantom{\rule{0ex}{0ex}}⇒z=36\left(\mathrm{cos}\frac{\pi }{2}+i\mathrm{sin}\frac{\pi }{2}\right)$
Now square roots of z is,

Hence the two square roots are,