# Finding radius for De Moivre's Theorem

• December 9th 2009, 05:48 AM
thekrown
Finding radius for De Moivre's Theorem
My question has (square root 2 / 2 - i square root 2 / 2)^20 and I must express this in the form a+bi.

I was able to find the angle of 315 degrees. Now I must find the radius, is it 1?

I've divided everything by square root of 2 since that is what it can simplify to, and then used the answer and used the formula r= square root of a^2+b^2. I got "1". Can you check if this is the radius?
• December 9th 2009, 07:07 AM
Quote:

Originally Posted by thekrown
My question has (square root 2 / 2 - i square root 2 / 2)^20 and I must express this in the form a+bi.

I was able to find the angle of 315 degrees. Now I must find the radius, is it 1?

I've divided everything by square root of 2 since that is what it can simplify to, and then used the answer and used the formula r= square root of a^2+b^2. I got "1". Can you check if this is the radius?

HI

$(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)^{20}$

$=(\frac{\sqrt{2}}{2})^{20}(1-i)^{20}$

$=\frac{2^{10}}{2^{20}}(-2^{10})$

$=1$

$(1-i)^2=-2i$

$
(1-i)^{20}=(-2i)^{20}=-2^{10}
$

so the magnitude would be 1 .