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arunma
Sep16-04, 07:39 PM
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
Sep16-04, 11:41 PM
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
Sep17-04, 04:44 AM
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
Sep18-04, 10:01 AM
Thanks arildno! I'll try your method; it looks familiar because the prof also mentioned something about a ± popping up in the problem.

arildno
Sep18-04, 10:06 AM
Good luck, arunma!
However, if you've learnt about complex exponentials, it is a quite instructive additional exercise to discover the equivalence of Tide's and my own approach..