Solve Complex Equation: 2x + 2iy + ix - y - 1 = 0

[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?

 

[Fermat]

You have reduced it correctly.

Your expression becomes,

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

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.

 

[cscott]

Thanks a lot, I got the answer now.

 

[Tide]

Your equation is equivalent to

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

so

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

or

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

which, I think, is a little more straightforward.

 

[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 truly work it out?

 

[Tide]

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

 

[cscott]

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

Not yet, no.

 

[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.

 

[cscott]

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.

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

Anyway, I got the answer. Thanks!