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## #1 2006-01-09 04:56:39

Peter T
Guest

### Help with evil complex numbers!

Question:
Find z such that
xx* + 3(z-z*) = 13 + 12i

can someone please tell me how to begin to solve this.

i realize that z* is the complex conjigate of z, but i just don't know where to start.

Thanks

## #2 2006-01-09 06:45:34

Ricky
Moderator

Offline

### Re: Help with evil complex numbers!

First, start by doing a few quick substitutions:

x   = a + bi (where a and b are real)
x* = a - bi

z   = c + di (where c and d are real)
z* = c - di

xx* = (a + bi)(a - bi) = a^2 + b^2
z - z* = c + di - (c - di) = 2di

So:

a^2 + b^2 + 3(2di) = 13 + 12i

Since i is the only possible imaginary value (all other variables must be real):

a^2 + b^2 = 13 and 6di = 12i

di = 2i, and since d must be real, d = 2

So z = c + 2i, where c is any real number.  The simpilest solution is z = 2i (c = 0)

Last edited by Ricky (2006-01-09 06:46:50)

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

## #3 2006-01-09 07:29:54

Peter T
Guest

### Re: Help with evil complex numbers!

thanks, i sort of understood what u did. ill be going along to my math support class to understand that fully.

## #4 2006-01-09 10:14:47

Ricky
Moderator

Offline

### Re: Help with evil complex numbers!

The reason why I did that is because x and x*, and z and z*, have a lot of the same terms in them, which you can take advantage of if you put them in full complex form.

From there, it's just plugging them in and trying to solve for c and d.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."