An alternate approach would be to rewrite the right side:
Here's what I've done so far.
Using the second equation:
Substituting into the first equation:
And that is where I'm stuck, I thought of using a horner scheme but still didn't make it
P.S: Could someone indicate how to correctly set the alignment of the last equation ?
You're fine - you just need to know a useful little trick to continue. Some 3-term polynomials are actually quadratics in disguise.
is , so it's "really" . Solving (I'll just factor it):
, so or .
When you start with z = x+iy, you're saying x and y are real numbers. Thus you can ignore the imaginary solutions for y, leaving only y = 3 and y = -3.
This trick can be useful. Similar examples: , and so forth.
-----
By the way, your approach is completely fine, but can also do it by putting everything into polar form (this approach works better with higher powers than quadratics):
There's an ERROR somewhere in what follows, but the approach is correct:
Let , where . Then
, where k is any integer.
Then , so or .
Note that , so
.
Thus or .
Thus or .
Thus or .
Thus or .
Triangle is , so .
Thus or .
Thus or .
Ugg - I blundered in here somewhere. It should be 1 + 3i and -1 - 3i. But the approach is correct, and good to know.