Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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

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.

share|cite|improve this answer
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 $$ \sqrt[4]{a^2+b^2}\left(\cos\frac\theta2 + i\sin\frac\theta2\right). $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.