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  • What is the polar form of  <img alt="\\\mathrm{z=\frac{4}{\sqrt{2}}-i\frac{3}{\sqrt{2}}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5C%5C%5Cmathrm%7Bz%3D%5Cfrac%7B4%

What is the polar form of  \\\mathrm{z=\frac{4}{\sqrt{2}}-i\frac{3}{\sqrt{2}}}  ?

Option: 1

\mathrm{\frac{5}{\sqrt{2}}\left ( -\cos\frac{3}{5} + i \sin\frac{4}{5}\right )}


Option: 2

\mathrm{\frac{5}{\sqrt{2}}\left ( \cos\frac{-3}{5} + i \sin\frac{4}{5}\right )}


Option: 3

\mathrm{\frac{5}{\sqrt{2}}\left ( \cos\frac{4}{5} - i \sin\frac{3}{5}\right )}


Option: 4

\mathrm{\frac{5}{\sqrt{2}}\left ( \cos\frac{4}{5} + i \sin\frac{-3}{5}\right )}


Answers (1)

best_answer

As we have learnt in

Polar form of complex numbers -

 

polar coordinate

\\\mathrm{z=x+iy}\\\mathrm{we\;can\;rewrite}\\\mathrm{z=\sqrt{x^2+y^2}\left ( \frac{x}{\sqrt{x^2+y^2}}+i\frac{y}{\sqrt{x^2+y^2}} \right )}\\\mathrm{Replace,\sqrt{x^2+y^2}=|z|,\;\;\;\sin\theta=\frac{y}{\sqrt{x^2+y^2}},\;\;and\;\;\cos\theta=\frac{x}{\sqrt{x^2+y^2}}}\

z = |z|(\cos\theta+\mathit{i}\sin\theta)

 this form is called polar form with \theta = principal value of arg(z) and r = |z|.

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We need to find |z| and argument of the z


\\\mathrm{arg(z)=\tan^{-1}\frac{\frac{-3}{\sqrt{2}}}{\frac{4}{\sqrt{2}}}=\tan^{-1}\frac{-3}{4}} \\\mathrm{\left | z \right | = \sqrt{\frac{16}{2}+\frac{9}{2}}=\frac{5}{\sqrt{2}}=r} \\\mathrm{So \;now \;in\; polar\; form \;z } =\mathrm{\left | Z \right |\arg(z)} \\\mathrm{\frac{5}{\sqrt{2}}\left ( \cos\frac{4}{5} + i \sin\frac{-3}{5}\right )}

Correct option is (d)

Posted by

Irshad Anwar

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