Complex Analysis: Nonlinear system

[arunma]

Here's a problem I ran into in complex analysis. Given z = x + iy and w = u + iv, I need to find all w such that w² = z. It reduces to solving this system:

x = u² - v²
y = 2uv

My professor mentioned that we should try to deal with the problem in at least two cases: y = 0, and y does not equal 0 (perhaps the second case should also be split up into a few cases). But beyond the y = 0 case, I'm stuck. Can anyone help?

 

[Tide]

Here's a problem I ran into in complex analysis. Given z = x + iy and w = u + iv, I need to find all w such that w² = z. It reduces to solving this system:

x = u² - v²
y = 2uv

My professor mentioned that we should try to deal with the problem in at least two cases: y = 0, and y does not equal 0 (perhaps the second case should also be split up into a few cases). But beyond the y = 0 case, I'm stuck. Can anyone help?

Why don't you try writing the complex numbers in complex exponential form form?

 

[arildno]

The second equation implies:
$$y^{2}=4u^{2}v^{2}$$
which is equivalent to (when substituting from the first equation):
$$4(v^{2})^{2}+4x(v^{2})-y^{2}=0$$
Or:
$$v^{2}=\frac{-4x\pm\sqrt{16x^{2}+16y^{2}}}{8}=\frac{-x\pm{r}}{2},r=\sqrt{x^{2}+y^{2}}$$
Comments:
1. Clearly, only non-negative values are acceptable for $$v^{2}$$
This means that we, for all x, have: $$v^{2}=\frac{r-x}{2}$$
2. Even more important, the equations:
$$y=2uv$$,$$y^{2}=4u^{2}v^{2}$$
are not equivalent; hence your answers may contain false solutions; ie, you must substitute what you get into your original system in order to determine the actual solutions.

 

[arunma]

Thanks arildno! I'll try your method; it looks familiar because the prof also mentioned something about a ± popping up in the problem.

 

[arildno]

Good luck, arunma!
However, if you've learned about complex exponentials, it is a quite instructive additional exercise to discover the equivalence of Tide's and my own approach..