Question

if x+iy=(1- i√3)^100,find x,y.

Answer 1

\[x+iy=(1- i√3)^{100}\]

Answer 2

Express \(1-i\sqrt3\) in the form of \(r(\cos\theta+i\sin\theta)\) first

Answer 3

how?

Answer 4

Any complex number in the form of \(a+bi\) can be expressed in the form of \(r(\cos\theta+i\sin\theta)\), where \(r=\sqrt{a^2+b^2}\) and \(\theta=\tan^{-1}\dfrac ba\)

Answer 5

For example, \(6+2i=\sqrt{6^2+2^2}(\cos(\tan^{-1}\dfrac26)+i\sin(\tan^{-1}\dfrac26))\)

Answer 6

I get : 2^99(1-√3i)^100

Answer 7

then?

Answer 8

@kc_kennylau

Answer 9

You forgot to change the things inside to cosine and sine?

Answer 10

no no, after solving cos and sin i get that..

Answer 11

For example, \(1-i=\sqrt{1^2+1^2}(\cos(\tan^{-1}\frac{-1}1)+\sin(\tan^{-1}\frac{-1}1))=\sqrt2(\cos135^\circ+i\sin135^\circ)\)

Answer 12

Don't solve the cosine and sine

Answer 13

@kc_kennylau : don't give examples please?

Here, r=2, and theta= -pi/3

Answer 14

there you get: 2(cos(-pi/3)+sin(-pi/3)) = 2(1/2-√3/2)

Answer 15

sorry, = [2(1/2 - √3/2)] ^100

Answer 16

= 2^99 (1-√3)^100 = 2^99(x+√3i)

Answer 17

So \((1-\sqrt3)^{100}=\{2[\cos(-\dfrac\pi3)+\sin(-\dfrac\pi3)]\}^{100}\)

Answer 18

Don't solve the cosine and sine

Answer 19

Or you'd have turned them into cosine and sine and then turned it back

Answer 20

Correction: \((1-\sqrt3)^{100}=\{2[\cos(-\dfrac\pi3)+i\sin(-\dfrac\pi3)]\}^{100}\)

Answer 21

then what do i do?

Answer 22

\[=2^{100}[\cos(-\frac\pi3)+\sin(-\frac\pi3)]^{100}\]

Answer 23

Theorem: \(\displaystyle\Large(\cos\theta+i\sin\theta)^n=\cos(n\theta)+i\sin(n\theta)\)

Answer 24

Correction: \(\displaystyle=2^{100}[\cos(-\frac\pi3)+i\sin(-\frac\pi3)]^{100}\)

Answer 25

Apply this theorem

Answer 26

\[\begin{array}{rcl}
(1-\sqrt3)^{100}&=&\{2[\cos(-\dfrac\pi3)+i\sin(-\dfrac\pi3)]\}^{100}\\
&=&2^{100}[\cos(-\frac\pi3)+i\sin(-\frac\pi3)]^{100}\\
&=&2^{100}[\cos(-\frac{100\pi}3)+i\sin(-\frac{100\pi}3)]\\
&=&2^{100}[\cos(-\frac{4\pi}3)+i\sin(-\frac{4\pi}3)]\\
&=&\mbox{now solve the cosine and sine}
\end{array}\]

Answer 27

AWESOME. I get : 2^99(-1 + √3i) == (which is the answer) :D

Answer 28

Glad you got your answer :)

Answer 29

Are you sure that you understood every step?

Answer 30

Yep :)