I have no clue how to do this one...

 

let a= -1+i, b=-2-i, solve for z. b*=complex conjugate of b

 

z-b*=Im(a^2/(2b+i))

It's a simple case of substituting in the values for a and b.

 

[math]z-\bar{b}=\Im(\frac{a^2}{2b+i})[/math]

 

[math]z = \Im(\frac{(-1+i)^2}{-4-i}) -2+i[/math]

 

Then to get a real denominator, multiply both the numerator and denominator by the conjugate of the current denominator.* (think difference of two squares to check that this always works)

 

[math]z = \Im(\frac{(-1+i)^2 (-4+i)}{17}) -2+i[/math]

 

Some mindless algebra follows (do it yourself)...

 

[math]z = \Im(\frac{-2-8i}{17}) -2+i[/math]

 

Etcetera... (not going to write down the final answer because this is obviously homework help, also you'll need to fill in a lot of the gaps.)

 

*That is, for [imath]\frac{p}{q}[/imath] with complex p and q, multiplying it by [imath]1=\frac{\bar{q}}{\bar{q}}[/imath] will allow you to separate the real and imaginary parts.