Question

Solve this complex numbers questions.

Also please check my other questions for answering.​

Answer 1

Answer:

option D

Step-by-step explanation:

because Cos( teeta )+ i ( Sin (teeta)) in which power 3 and 4 are subtracted and became 1

Answer 2

Given Question :-

Value of

$\rm :\longmapsto\:\dfrac{(cos\theta + isin\theta)^{4} }{ {(cos\theta -isin\theta) }^{3} }$

$\sf \: \: \: \: \: (a) \: \: cos5\theta + isin5\theta$

$\sf \: \: \: \: \: (b) \: \: cos7\theta + isin7\theta$

$\sf \: \: \: \: \: (c) \: \: cos4\theta + isin4\theta$

$\sf \: \: \: \: \: (d) \: \: cos\theta + isin\theta$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\dfrac{(cos\theta + isin\theta)^{4} }{ {(cos\theta -isin\theta) }^{3} }$

can be rewritten as

$\rm \: = \:\dfrac{(cos\theta + isin\theta)^{4} }{ {(cos( - \theta) + isin( - \theta)) }^{3} }$

We know, that

By Euler Form

$\rm :\longmapsto\:\boxed{ \tt{ \: cos\theta + isin\theta = {e}^{i\theta} }}$

So, using this, we get

$\rm \: = \:\dfrac{ {( {e}^{i\theta} )}^{4} }{ {( {e}^{ - i\theta)} }^{3} }$

$\rm \: = \:\dfrac{ {e}^{i4\theta} }{ {e}^{ - i3\theta} }$

$\rm \: = \: {e}^{4i\theta - ( - 3i\theta)}$

$\rm \: = \: {e}^{4i\theta + 3i\theta}$

$\rm \: = \: {e}^{i7\theta}$

$\rm \: = \: {e}^{i(7\theta)}$

$\rm \: = \:cos7\theta + i \: sin7\theta$

Hence,

$\rm :\longmapsto\:\boxed{ \tt{ \: \dfrac{(cos\theta + isin\theta)^{4} }{ {(cos\theta -isin\theta) }^{3} } = cos7\theta + isin7\theta \: }}$

So,

• Option (b) is Correct