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
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
Hello, ArTiCK!
Ratinalize the "inside" first . . .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: .$\displaystyle (-i)^7 \;=\; i$
$\displaystyle \left(\frac{1-i\sqrt{3}}{\sqrt{3}+i}\right)^7 =
$ $\displaystyle \left[ \frac{2 \, \text{cis} \left( -\frac{\pi}{3}\right) }{2 \, \text{cis} \left( \frac{\pi}{6}\right)} \right]^7 = \left[ \text{cis} \left( -\frac{\pi}{3} - \frac{\pi}{6} \right)\right]^7 = \left[ \text{cis} \left( -\frac{\pi}{2} \right)\right]^7 = (-i)^7$.