-1
$\begingroup$

Question: Prove that if $z=x+i y$ then $w=u+iv$ where $w^{2}=z$, then $w$ is the square root of $z$ and is donated by $\sqrt{z}$

I have an exam next week and I am trying to solve some past exam questions. This is an old exam question. The question seems obvious but I am not sure how to prove it. Is the solution to define $w$ in terms of $x$ and $y$ then show that $w^{2}=z$?

  • 0
    I'm not quite sure what you are asking. Are you asking about the existence of $w$? Or are we supposed to prove something about $w$? Or $u,v$? – 2017-01-04
  • 0
    @SimpleArt, This is exactly what the question asks I am confused by it too. – 2017-01-04
  • 3
    Are you quoting the question _exactly as presented to you_ including _all parts_ of the question? It's not at all clear what it asks you to prove, and writing "$w^2=z$ is the square root of $z$" appears to be a pretty horrible typo. – 2017-01-04
  • 0
    @HenningMakholm, this is exact question. I think it means $w$ is square root to $z$. – 2017-01-04
  • 1
    I'm skeptical that this is an entire question. Can you upload a scan/photo of the entire page that contains this question? – 2017-01-04
  • 0
    @HenningMakholm: please be charitable to $\sqrt{z}$ who was kind enough to donate $w$ to us $\ddot{\smile}$. – 2017-01-04
  • 0
    It seems we will never know the meaning of the question (hasty acceptance, no explanation about the text of the question...). – 2017-01-04

1 Answers 1

3

Substitute your values in:

$$(u+iv)^2=x+iy$$

$$u^2-v^2+2iuv=x+iy$$

Since it is assumed that when complex numbers are written in this form that the coefficients are real, thus, we can equate parts:

$$u^2-v^2=x\\2uv=y$$

Solving for $u$ in the latter equation, we have

$$u=\frac y{2v}$$

Substitute it into the other equation:

$$\left(\frac y{2v}\right)^2-v^2=x$$

$$y^2-4v^4=4xv^2$$

$$0=4v^4+4xv^2-y^2$$

This is a polynomial of $v^2$, and we apply the quadratic formula:

$$v^2=\frac{-x\pm\sqrt{x^2+y^2}}2$$

$$v=\pm\sqrt{\frac{-x\pm\sqrt{x^2+y^2}}2}$$

Similar, $v$ can be solved for. One chooses the inner $\pm$ to be $+$ so that the square root is positive, or else $v$ would be complex:

$$v=\pm\sqrt{\frac{-x+\sqrt{x^2+y^2}}2}$$

I leave you to solve for $u$, with the note that the outer $\pm$ stays since

$$(\pm w)^2=z$$

  • 0
    Starting at $$y^2-2v^4=2xv^2$$ one should read $$y^2-4v^4=4xv^2$$ – 2017-01-04
  • 0
    @Did My bad :D Thanks for the proof read – 2017-01-04
  • 0
    That was easy, the square roots $\sqrt{x^2+2y^2}$ being impossible. – 2017-01-04
  • 0
    @Did I can't really blame them if they did check and missed the errors, for they themselves asking the question might mean they are prone to such errors themselves. And my answer has been sitting here long enough for the OP to go through this. – 2017-01-04
  • 0
    "...has **not** been sitting here long enough for the OP to go through this", you mean? *Which is exactly the problem.* :-) – 2017-01-04
  • 0
    @Did, I did check. I know what is wrong with the answer but I can solve it myself now since I know how to go about it. – 2017-01-04
  • 0
    @gbd What? And commenting on the mistake was not an option because? – 2017-01-04
  • 0
    @Did, I think you have already commented about the mistake before I noticed it. – 2017-01-04
  • 0
    @gbd Back to square one then: you accepted an answer without checking it. Why the hurry? – 2017-01-04