Hi all,

I am having trouble getting the answer to:

[(1-sqrt(3)i)/ (sqrt(3) + i)]^7

where i = complex number

Thanks in advance,

ArTiCk

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- Jun 11th 2008, 06:03 PMArTiCKcomplex numbers
Hi all,

I am having trouble getting the answer to:

[(1-sqrt(3)i)/ (sqrt(3) + i)]^7

where i = complex number

Thanks in advance,

ArTiCk - Jun 11th 2008, 06:28 PMmr fantastic
- Jun 11th 2008, 06:38 PMArTiCK
I have already converted the numbers into polar form...

i have it = e^(-7i*pi/3)/ [e^(7i*pi/6)]

The trouble i have is that i tried converting e^(-7i*pi/3) into Cartesian form as well as the denominator... but i am having trouble with the nominator. What i did was:

e^(-7i*pi/3) = cos(-7*pi/3) + i*sin(-7*pi/3) = cos (-2*pi - pi/3) + i*sin(-2*pi -pi/3) = cos(-pi/3) + i*sin(-pi/3)

I think it is there where i have gone wrong, help would be appreciated

Thanks, ArTiCk - Jun 11th 2008, 06:52 PMSoroban
Hello, ArTiCK!

Quote:

Simplify: .$\displaystyle \left(\frac{1-i\sqrt{3}}{\sqrt{3}+i}\right)^7$

. . $\displaystyle \frac{1-i\sqrt{3}}{\sqrt{3} + i}\cdot\frac{\sqrt{3}-i}{\sqrt{3}-i} \;=\;\frac{\sqrt{3} - i - 3i +i^2\sqrt{3}}{3-i^2} \;=\;\frac{\sqrt{3} -4i-\sqrt{3}}{3+1} \;=\;\frac{-4i}{4} \;=\;-i$

So we have: .$\displaystyle (-i)^7 \;=\; i$

- Jun 11th 2008, 07:08 PMmr fantastic
$\displaystyle \left(\frac{1-i\sqrt{3}}{\sqrt{3}+i}\right)^7 =

$ $\displaystyle \left[ \frac{2 \, \text{cis} \left( -\frac{\pi}{3}\right) }{2 \, \text{cis} \left( \frac{\pi}{6}\right)} \right]^7 = \left[ \text{cis} \left( -\frac{\pi}{3} - \frac{\pi}{6} \right)\right]^7 = \left[ \text{cis} \left( -\frac{\pi}{2} \right)\right]^7 = (-i)^7$. - Jun 11th 2008, 07:34 PMSoroban