Martin Hart
2022-10-22
Answered

Find the two square roots of 36i.

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getrdone07tl

Answered 2022-10-23
Author has **23** answers

$z=36i\phantom{\rule{0ex}{0ex}}=0+36i$

Now,

$r=\sqrt{{0}^{2}+(36{)}^{2}}\phantom{\rule{0ex}{0ex}}=36$

&

$\theta ={\mathrm{tan}}^{-1}(\frac{x}{y})\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}(\frac{36}{0})\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}(\mathrm{\infty})\phantom{\rule{0ex}{0ex}}=\frac{\pi}{2}$

Now the polar form is,

$z=r(\mathrm{cos}\theta +i\mathrm{sin}\theta )\phantom{\rule{0ex}{0ex}}\Rightarrow z=36(\mathrm{cos}\frac{\pi}{2}+i\mathrm{sin}\frac{\pi}{2})$

Now square roots of z is,

$\sqrt{z}=[36(\mathrm{cos}\frac{\pi}{2}+i\mathrm{sin}\frac{\pi}{2}){]}^{\frac{1}{2}}\phantom{\rule{0ex}{0ex}}=6\pm (\mathrm{cos}\frac{\pi}{2}+i\mathrm{sin}\frac{\pi}{2}{)}^{\frac{1}{2}}\phantom{\rule{0ex}{0ex}}=6\pm (\mathrm{cos}(\frac{1}{2}\cdot \frac{\pi}{2})+i\mathrm{sin}(\frac{1}{2}\cdot \frac{\pi}{2}))\text{}\text{}\text{}[\therefore (\mathrm{cos}\theta +i\mathrm{sin}\theta {)}^{n}=\mathrm{cos}n\theta +i\mathrm{sin}n\theta ]\phantom{\rule{0ex}{0ex}}=6\pm (\mathrm{cos}(\frac{\pi}{4})+i\mathrm{sin}(\frac{\pi}{4}))\phantom{\rule{0ex}{0ex}}=6\pm (\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})\phantom{\rule{0ex}{0ex}}=\pm (3\sqrt{2}+3i\sqrt{2})$

Hence the two square roots are,

$-3\sqrt{2}-3i\sqrt{2}\text{}\text{}\text{}3\sqrt{2}+3i\sqrt{2}$

Now,

$r=\sqrt{{0}^{2}+(36{)}^{2}}\phantom{\rule{0ex}{0ex}}=36$

&

$\theta ={\mathrm{tan}}^{-1}(\frac{x}{y})\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}(\frac{36}{0})\phantom{\rule{0ex}{0ex}}={\mathrm{tan}}^{-1}(\mathrm{\infty})\phantom{\rule{0ex}{0ex}}=\frac{\pi}{2}$

Now the polar form is,

$z=r(\mathrm{cos}\theta +i\mathrm{sin}\theta )\phantom{\rule{0ex}{0ex}}\Rightarrow z=36(\mathrm{cos}\frac{\pi}{2}+i\mathrm{sin}\frac{\pi}{2})$

Now square roots of z is,

$\sqrt{z}=[36(\mathrm{cos}\frac{\pi}{2}+i\mathrm{sin}\frac{\pi}{2}){]}^{\frac{1}{2}}\phantom{\rule{0ex}{0ex}}=6\pm (\mathrm{cos}\frac{\pi}{2}+i\mathrm{sin}\frac{\pi}{2}{)}^{\frac{1}{2}}\phantom{\rule{0ex}{0ex}}=6\pm (\mathrm{cos}(\frac{1}{2}\cdot \frac{\pi}{2})+i\mathrm{sin}(\frac{1}{2}\cdot \frac{\pi}{2}))\text{}\text{}\text{}[\therefore (\mathrm{cos}\theta +i\mathrm{sin}\theta {)}^{n}=\mathrm{cos}n\theta +i\mathrm{sin}n\theta ]\phantom{\rule{0ex}{0ex}}=6\pm (\mathrm{cos}(\frac{\pi}{4})+i\mathrm{sin}(\frac{\pi}{4}))\phantom{\rule{0ex}{0ex}}=6\pm (\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2})\phantom{\rule{0ex}{0ex}}=\pm (3\sqrt{2}+3i\sqrt{2})$

Hence the two square roots are,

$-3\sqrt{2}-3i\sqrt{2}\text{}\text{}\text{}3\sqrt{2}+3i\sqrt{2}$

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