# Complex Equation

1. Dec 15, 2005

### cscott

Solve the following equations for the real variables x and y.
(b) $$\left ( \frac{1 + i}{1 - i} \right )^2 + \frac{1}{x + iy} = 1 + i$$
I reduced this to
$$2x + 2iy + ix - y - 1 = 0$$
but I cannot get any further. Have I reduced it correctly?

Last edited: Dec 15, 2005
2. Dec 15, 2005

### Fermat

You have reduced it correctly.

$$(2x-y-1) + i(2y+x) = 0$$

The rhs is zero, so the lhs must be zero. And the only way that can happen is for both the real component and the imaginary component to be equal to zero.

3. Dec 15, 2005

### cscott

Thanks a lot, I got the answer now.

4. Dec 15, 2005

### Tide

$$-1 + \frac{1}{x + iy} = 1 + i$$

so

$$\frac {1}{x + iy} = 2 + i$$

or

$$x + iy = \frac {1}{2 + i}$$

which, I think, is a little more straightforward.

5. Dec 15, 2005

### cscott

I've got another:

Find the square root

$$(x + iy)^2 = -3 + 4i$$

I think I'm supposed to do some hand-waving here, but I would I truely work it out?

6. Dec 15, 2005

### Tide

Have you learned to express complex numbers in polar form yet?

7. Dec 16, 2005

### cscott

Not yet, no.

8. Dec 16, 2005

### Tide

Okay. In that case, just expand $(x+iy)^2$ then equate real and imaginary parts on the left and right sides of the equation. This gives you two equations and two unknowns (x and y) to solve for. It's fairly straightforward. Let us know what you come up with.

9. Dec 16, 2005

### cscott

Doh, why didn't I think of that!?

Anyway, I got the answer. Thanks!