Solving the Equation for a Complex Number

[adam640]

Homework Statement

Solve the equation \overline{y} (y - 2) = 2\overline{y} + 15 - 8i for complex number y

Homework Equations

\overline{y}y = a² + b²

The Attempt at a Solution

a²+b²+8i-4ib-15=0
a²+b(b-4i)+8i-15=0

Pretty clueless where to go from here? Or if I've even gone in the right direction. Any help would be greatly appreciated.
Thanks,
Adam

 

[LCKurtz]

Check your arithmetic and when you are sure it is correct set it equal to 0 + 0i and equate real and imaginary parts.

 

[adam640]

I have set the equation to = 0 but I don't understand how I can set to 0i though? I'm sure about the maths so far.

 

[eumyang]

0 = 0 + 0i

Say you have a simple complex equation:
a + 3 + 2bi = 0

This is the same as saying
(a + 3) + 2bi = 0 + 0i

Equate the real and imaginary coefficients:
a + 3 = 0 => a = -3
2b = 0 => b = 0

 

[adam640]

Thanks for the help, from what you've said I've gathered...

From a²+b(b-4i)+8i-15=0

0=0+0i
Therefore
From a²+b(b-4i)+8i-15=0+0i
a²-15=0
=> a = [STRIKE]\sqrt{15}[/STRIKE]
b²-4b-8 = 0
b is solved with the quadratic?

Is this correct so far?

 

[LCKurtz]

Thanks for the help, from what you've said I've gathered...

From a²+b(b-4i)+8i-15=0

That equation isn't correct. Check your work

0=0+0i
Therefore
From a²+b(b-4i)+8i-15=0+0i
a²-15=0
=> a = [STRIKE]\sqrt{15}[/STRIKE]
b²-4b-8 = 0

And that last step is wrong too. b(b-4i) isn't (b²-4b)i.

b is solved with the quadratic?

Is this correct so far?