Complex Numbers - Find the proof for "z"

Question

Hi I have a question on complex numbers, where I'm not sure how to go about dealing with it.

This is what I need to do:

Let $z=a+ib$, be a complex number. Show that a square root of $z$, is given by the expression: $$w=\sqrt{(\mid z\mid +a)/2} + i\sigma\sqrt{(\mid z\mid -a)/2}$$ where $\sigma=1$ if $b≥0$ and $\sigma=-1$ if $b<0$. Do this by verifying that $w^2=z$.

Do i need to go about this by starting with $w=\sqrt z=\sqrt{a+ib}$ and then manipulating the $a+ib$ part, or how would I do this?

Thanks.

Answer 1

Take $w=\sqrt{(\mid z\mid +a)/2} + i\sigma\sqrt{(\mid z\mid -a)/2}$ and compute (with this expression) $w^2$. If you do it right, you should get $w^2=z$.

Answer 2

No. Understand that what you suggest makes no sense, because $\sqrt{z}$ makes no sense when $z$ is a complex number. There are more than one square root of $z$, and that is the meaning of "...show that "a" square root of..." as opposed to "the" square root.

What you have to do is to show that $$w^2=z$$ as in fact suggested at the end of the question that you posted.

Answer 3

Let $\sqrt z=u+iv$. Then by squaring and identifying,

$$\begin{cases}u^2-v^2=a,\\2uv=b.\end{cases}$$

Multiply the first equation by $u^2$ and eliminate $v$ with the second.

$$u^2-au^2-\frac{b^2}4=0.$$

The discriminant is $a^2+b^2=|z|^2$ and the solution in $u$ then $v$

$$\begin{cases}u=\pm\sqrt{\dfrac{|z|+a}2},\\v=\dfrac b{2u}.\end{cases}$$ (The solution with $a-|z|$ must be rejected, as $u$ is real.)

Notice that after simplification, $v$ can be written

$$v=\pm\sqrt{\frac{|z|-a}2}.$$

Finally, to assign the signs, notice that $uv$ has the sign of $b$, let $\sigma$, and you remain free to choose the branch that suits you.

$$\begin{cases}u=\sqrt{\dfrac{|z|+a}2},\\v=\sigma\sqrt{\dfrac{|z|-a}2}.\end{cases}$$