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Can anyone help me with this proof? I am not sure how to exactly go about this using just variables such as a + bi ?

Thanks in advance!

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Perhaps let W=a+bi, and Z=c+di. Then it is just a matter of finding c and d. So, determine the real and imaginary part of 1/Z, and add it to Z to get a complex number in terms of c and d. Now you should have an expression from which you can solve for x and d in terms of a and b... –  Peter Grill Mar 16 '12 at 1:50

2 Answers 2

Hint $\:\mathbb C$ algebraically closed $\rm\Rightarrow\: f(z) = z^2 - w\: z + 1\:$ has a root $\rm\:r\in\mathbb C.\:$ $\rm\:r\ne 0\:$ since $\rm\:f(0) = 1.$

To determine the root simply apply the quadratic formula, and see this post on calculating square roots of complex numbers.

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Thanks for your help Bill. I think I got it. Thanks for pointing me in the right direction! –  JackReacher Mar 16 '12 at 5:12
    
You really don't need anything as sophisticated as knowing that $\mathbb{C}$ is algebraically closed in order to know that you can solve quadratic equations within $\mathbb{C}$, do you? –  Michael Hardy Mar 18 '12 at 23:45
    
@Michael See the link. –  Bill Dubuque Mar 18 '12 at 23:53
    
@MichaelHardy: But then you can solve the same problem for many more cases. –  Najib Idrissi Mar 20 '12 at 19:33

The angle from the positive real axis to the ray from $0$ through $a+bi$ is $\theta=\arctan(b/a)$ if $a\ge0$ and $\theta=\pi+\arctan(b/a)$ if $a<0$. For the square root, you want just half that angle.

The absolute value $|a+bi|$ is $\sqrt{a^2+b^2}$. For the square root of $a+bi$, you want the square root of that, so that's $\sqrt[4]{a^2+b^2}$.

So you get $$ \sqrt[4]{a^2+b^2}\left(\cos\frac\theta2 + i\sin\frac\theta2\right). $$

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