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\text{z}=\frac{16}{1+i\sqrt{3}}   , its Euler form is?

Option: 1

-8e^{\frac{\pi }{3}i}


Option: 2

\sqrt{8}e^{-\frac{\pi }{3}i}


Option: 3

8e^{\frac{\pi }{3}i}


Option: 4

8e^{-\frac{\pi }{3}i}


Answers (1)

best_answer

As we have learnt in

 

 

Euler form of complex number -

The polar form of complex number z=r(cos ? + i sin ?), in Euler form (cos ? + i sin ?) part of the polar form of complex numbers is represented by e .

 

 e = cos? + isin? and  e-iΘ = cos? - isin?

Euler forms make algebra very simple for complex numbers. Any complex number can be expressed as

 

\\\mathrm{z=x+iy\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\(Cartesian\;form)}\\\mathrm{z=|z(\cos\theta+i\sin\theta)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(Polar\;form)}\\\mathrm{z=|z|e^{i\theta}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\(Eular's\;form)}

 

 

we simplify z, for that we normalize the denominator 

\\\mathrm{z=\frac{16}{1+i\sqrt{3}}\cdot \frac{1-i\sqrt{3}}{1-i\sqrt{3}}=4(1-i\sqrt{3})}

Now we see it lies in the 4th quadrant, so the argument is going to be -ve.

 First we find r = |z| = 4·2=8

 

\\\mathrm{\theta = arg(z)=\tan^{-1}(-\sqrt{3})=\frac{-\pi}{3}} \\\mathrm{So \;euler\; form = re^{i\theta} = 8e^{-\frac{\pi }{3}i}}

 

Correct option is (d)

Posted by

avinash.dongre

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