Question

Use De Moivre's theorem to write the complex number in trigonometric form. (cospi/4 + isinpi/4)^3

Answer 1

Given complex number ,

$({ \cos( \frac{\pi}{4} ) + i \sin( \frac{\pi}{4} )) }^{3}$

to find :

write the complex number in trigonometric form by using DE movier theorem

solution:

De movier's theorem :

${( \cos(x) + i \sin(x)) }^{n} = ( \cos(nx) + i \sin(nx) )$

=>

$( \cos( \frac{\pi}{4} ) + i \sin( \frac{\pi}{4} ) ) ^{3} = \cos( \frac{3\pi}{4} ) + i \sin( \frac{3\pi}{4} )$

$\: \: \: = \cos(\pi - \frac{\pi}{4} ) + i \sin(\pi - \frac{\pi}{4} )$

${\cos(\pi - x ) = - \cos(x) }$

$( \sin(\pi - x) = \sin(x) )$

$= - \cos( \frac{\pi}{4} ) + i \sin( \frac{\pi}{4} )$

$= - \frac{1 }{ \sqrt{2} } + i \frac{1}{ \sqrt{2} }$

therefore ,

The complex number in trigonometric form :

$= - \frac{1 }{ \sqrt{2} } + i \frac{1}{ \sqrt{2} }$

Answer 2

Answer:

As per the data given in the above question

Given the complex number

$z_1 = (\cos\frac{\pi}{4} + i \sin\frac{\pi}{4})$

we know that according to De Moivre's Theorem

for a complex number,

$z = |z|(\cos\theta + i \sin\theta)$

$z^n = |z|^n(\cos n \theta + i \sin n \theta)$

we want to find the value of norm of $z_1$

applying the formula

$|z_1| = \sqrt{(\cos\frac{\pi}{4})^2 + (\sin\frac{\pi}{4})^2} = \sqrt{1} = 1$

so now according to De Moivre's Theorem,

we have $n=3$ and $\theta = \frac{\pi}{4}$

${z_1}^3 = (\cos\frac{\pi}{4} + i \sin\frac{\pi}{4})^3\\{z_1}^3 = {|z_1|}^3(\cos 3 \frac{\pi}{4} + i \sin3\frac{\pi}{4})\\{z_1}^3 = {1}^3({ - \frac{1}{\sqrt2}} + i {\frac{1}{\sqrt2}})\\ {z_1}^3 = ({ - \frac{1}{\sqrt2}} + i {\frac{1}{\sqrt2}})$

Hence the complex number in trigonometry form using De Moiver's theorem is $({ - \frac{1}{\sqrt2}} + i {\frac{1}{\sqrt2}})$ .