Need help with an imaginary number: polar form of 1 - (sqrt 3 - i)/2 ?

[Omgthehorror]

What is the polar form of 1 - (sqrt 3 - i)/2 ?

I understand the concept of finding r and the angle, but I seem to have hit a wall with this one. When I try solving it, I get r^2 = 2 - sqrt 3, but that is not a good looking number as far as I'm concerned. I need to express it in pi radians. Am I missing something ?

 

[Steven G]

What is the polar form of 1 - (sqrt 3 - i)/2 ?

I understand the concept of finding r and the angle, but I seem to have hit a wall with this one. When I try solving it, I get r^2 = 2 - sqrt 3, but that is not a good looking number as far as I'm concerned. I need to express it in pi radians. Am I missing something ?

Assume for a moment that 2 - sqrt 3 is a good looking number. What would the next step be?

 

[pka]

What is the polar form of 1 - (sqrt 3 - i)/2 ?

I understand the concept of finding r and the angle, but I seem to have hit a wall with this one. When I try solving it, I get r^2 = 2 - sqrt 3, but that is not a good looking number as far as I'm concerned. I need to express it in pi radians. Am I missing something ?

Always, Always, Always write a complex number in proper format ​\(\displaystyle Re(z)+Im(z)\bf{i}\).
So 1 - (sqrt 3 - i)/2 should be written as \(\displaystyle \left(\frac{2-\sqrt3}{2}+\frac{1}{2}\bf{i}\right)\)

Now to find the argument of the complex number:
Suppose that \(\displaystyle x\cdot y\ne 0 \) then
\(\displaystyle \arg(x + y\bf{i}) = \left\{ {\begin{array}{{rl}} {\arctan \left( {\frac{y}{x}} \right),}&{x > 0} \\ {\arctan \left( {\frac{y}{x}} \right) + \pi ,}&{x < 0\;\& \;y > 0} \\ {\arctan \left( {\frac{y}{x}} \right) - \pi ,}&{x < 0\;\& \;y < 0} \end{array}} \right. \)

This in this case \(\displaystyle \theta=\arctan\left(\frac{1}{2-\sqrt3}\right)\)

 

[Omgthehorror]

Thanks for the replies. I solved it with a calculator. I was put off by the fact that all of the previous problems were made so you got a cos and sin values that directly corresponded to the unit circle and this one I had to use a calculator to figure out the arccos.