0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ The question tells you exactly what to do, and that is to verify $w^2=z$... Reading the question usually helps a lot when you are trying to answer it... $\endgroup$ – 5xum Mar 20 '18 at 13:21
1
$\begingroup$

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 $w^2=z$.

$\endgroup$
1
$\begingroup$

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.

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ You are right that $\sqrt{z}$ is ambiguous for complex $z$. But the question uses the $\sqrt{}$ sign only for real arguments, so is correctly posed since it asks to verify a square root. Yes, the way to check is just to square $w$. $\endgroup$ – Ethan Bolker Mar 20 '18 at 12:09
  • $\begingroup$ @EthanBolker "the question uses the √ sign only for real arguments" no it doesn't, the OP was asking whether to start with $\sqrt{z}=\sqrt{a+ib}$ and manipulate this. I pointed out that this makes no sense. $\endgroup$ – Arnaud Mortier Mar 20 '18 at 12:24
  • $\begingroup$ OK I see. Sorry. $\endgroup$ – Ethan Bolker Mar 20 '18 at 12:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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