1. ## complex numbers

Hi all,

I am having trouble getting the answer to:
[(1-sqrt(3)i)/ (sqrt(3) + i)]^7

where i = complex number

ArTiCk

2. Originally Posted by ArTiCK
Hi all,

I am having trouble getting the answer to:
[(1-sqrt(3)i)/ (sqrt(3) + i)]^7

where i = complex number

ArTiCk
My advice is to first convert the numbers into polar form.

3. Originally Posted by mr fantastic
My advice is to first convert the numbers into polar form.
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

4. Hello, ArTiCK!

Simplify: . $\left(\frac{1-i\sqrt{3}}{\sqrt{3}+i}\right)^7$
Ratinalize the "inside" first . . .

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

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

5. Originally Posted by 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
$\left(\frac{1-i\sqrt{3}}{\sqrt{3}+i}\right)^7 =
$
$\left[ \frac{2 \, \text{cis} \left( -\frac{\pi}{3}\right) }{2 \, \text{cis} \left( \frac{\pi}{6}\right)} \right]^7 = \left[ \text{cis} \left( -\frac{\pi}{3} - \frac{\pi}{6} \right)\right]^7 = \left[ \text{cis} \left( -\frac{\pi}{2} \right)\right]^7 = (-i)^7$.

6. Hello, Mr. F !

Originally Posted by mr fantastic

$\left(\frac{1-i\sqrt{3}}{\sqrt{3}+i}\right)^7 =
$
$\left[ \frac{2 \, \text{cis} \left( -\frac{\pi}{3}\right) }{2 \, \text{cis} \left( \frac{\pi}{6}\right)} \right]^7 = \left[ \text{cis} \left( -\frac{\pi}{3} - \frac{\pi}{6} \right)\right]^7 = \left[ \text{cis} \left( -\frac{\pi}{2} \right)\right]^7 = (-i)^7$.

Lovely!