Question

(1-i)^5 using de moivres theorem.​

Answer 1

Answer:

$- 4 + 4i$

Step-by-step explanation:

To find ---->

$value \: of \: {( \: 1 \: - \: i \: )}^{5} \: by \: using \: de \: moivers \: theorem$

Concept used---->

1)

$\cos( - \alpha ) \: = \cos( \alpha ) \\ \sin( - \alpha ) \: = - \sin( \alpha )$

2)

${( \: cos \alpha + isin \alpha \: ) }^{n} = cos \: n \alpha + isin \: n \alpha$

3)

$\cos(\pi + \alpha ) \: = - \: cos \alpha \\ \sin(\pi + \alpha ) \: = - sin \alpha$

solution---->

${( \: 1 \: - \: i \: )}^{5}$

=$( \sqrt{2} \:\: ( \dfrac{1}{ \sqrt{2} } - \dfrac{1}{ \sqrt{2} }i \: ) \: ) ^{5}$

$= {( \: \sqrt{2} \: ) }^{5} \: { ( \: \dfrac{1}{ \sqrt{2} \: } \: - \dfrac{1}{ \sqrt{2} }i ) }^{5}$

$= 4 \sqrt{2} \: {( \: cos \dfrac{\pi}{4} \: - i \: sin \dfrac{\pi}{4} \: ) }^{5}$

$= 4 \sqrt{2} \: {( \: cos( - \frac{\pi}{4}) \: + \: i \: sin( - \frac{\pi}{4}) \: ) }^{5}$

$= 4 \sqrt{2} \: ( \: \cos( - \dfrac{5\pi}{4}) \: + \: i \: \sin( - \dfrac{5\pi}{4} ) \: )$

$= 4 \sqrt{2} \: ( \: cos \dfrac{5\pi}{4} \: - i \: sin \dfrac{5\pi}{4} \: )$

$= 4 \sqrt{2} \: ( \: \cos(\pi \: + \dfrac{\pi}{4} ) \: - i \: \sin( \: \pi \: + \dfrac{\pi}{4} ) \: )$

$= 4 \sqrt{2} \: ( \: - \cos( \dfrac{\pi}{4} ) \: + \: i \: \sin( \dfrac{\pi}{4} ) \: )$

$= 4 \sqrt{2} \: ( \: - \dfrac{1}{ \sqrt{2} } \: + \: i \: \dfrac{1}{ \sqrt{2} } \: )$

$= - \dfrac{4 \sqrt{2} }{ \sqrt{2} } \: + \: i \: \dfrac{4 \sqrt{2} }{ \sqrt{2} }$

$= \: - 4 \: + \: 4i$

Answer 2

$\huge\boxed{\fcolorbox{violet}{violet}{Answer}}$