1. ## Precalc- complex numbers

The system of equations
has two solutions
and in complex numbers. Find .

So far I replaced z with a+bi, did some algebra (squared both equations) and ended up with these equations:

a^2 + b^2 + (4-4a)i + 4b = 23

and

a^2 + b^2 - 16a + 10b + (80 - 10a - 16b )i = -1

Not sure where to go from here. Thanks!

2. ## Re: Precalc- complex numbers

This may help.

3. ## Re: Precalc- complex numbers

Originally Posted by geisha
The system of equations
has two solutions
and in complex numbers. Find .

So far I replaced z with a+bi, did some algebra (squared both equations) and ended up with these equations:

a^2 + b^2 + (4-4a)i + 4b = 23

and

a^2 + b^2 - 16a + 10b + (80 - 10a - 16b )i = -1

Not sure where to go from here. Thanks!

You don't really want to start by squaring. The | |'s in this case stand for the magnitude of the complex expression inside them. This isn't the same as the square of that expression.

$\left \mid a + i b \right \mid = \sqrt{a^2 + b^2}$

so for example in your first equation

$\left \mid z-2-2i \right \mid = \sqrt{23}$

let $z=a+bi$

$\left \mid (a-2)+(b-2)i \right \mid=\sqrt{(a-2)^2+(b-2)^2}=\sqrt{23}$

do the same thing for your second equation and then you can square both sides and solve the quadratic system for a and b.

4. ## Re: Precalc- complex numbers

Got it, thanks!!