This may sound like silly question, but I've made multiple attempts and can't figure out what do do with this question:
Find all solutions to z^{2} + 4z' + 4 = 0 where z is a complex number.
*z' being the conjugate
Ok I get that.
I was confused initially because I wasn't really sure how to go about the conjagate.
That being because it's the imaginary number that get's its sign swapped. So Wasn't sure if i could change the sign of a whole complex number.
Brilliant