Hello, It's me again, with yet another easy question that I can't answer 
I think I must've missed something In school along the track because I keep getting completely stuck. Id greatly appreciate it if someone could look through my working and tell me where I went wrong, or what I have to do next.
Here is the question:
Solve for the real numbers a and b:
\(\displaystyle (a+bi)^2 = 2 - 2 sqrt 3i\)
\(\displaystyle a^2 - b^2 - 2abi = 2 - 2 sqrt 3i\)
\(\displaystyle a^2 - b^2 = 2 sqrt 3i\)
\(\displaystyle 2ab = 2 sqrt 3b\)
Therefore... \(\displaystyle a = (2 sqrt 3b)/2\)
Substitute it in:
\(\displaystyle ((2 sqrt 3b)/2)^2 - b^2 = 2\)
\(\displaystyle (4 sqrt 9)/4b^2 - b^2 = 2\)
\(\displaystyle 12/4b^2 -b^2 = 2\)
\(\displaystyle 12 - b^2 = 8b^2\)
\(\displaystyle 0 = 8b^2 + b^2 - 12\)
\(\displaystyle 0 = 9b^2 - 12\)
Ok Stuck
I'd be very happy if someone could help.
Oh also, in the same question, which I did a different way, I got \(\displaystyle a^2 + 2a = 3\) ---it's obvious that a is one, but how do I get from there to an answer?
Thanks alot
-Adam
I think I must've missed something In school along the track because I keep getting completely stuck. Id greatly appreciate it if someone could look through my working and tell me where I went wrong, or what I have to do next.
Here is the question:
Solve for the real numbers a and b:
\(\displaystyle (a+bi)^2 = 2 - 2 sqrt 3i\)
\(\displaystyle a^2 - b^2 - 2abi = 2 - 2 sqrt 3i\)
\(\displaystyle a^2 - b^2 = 2 sqrt 3i\)
\(\displaystyle 2ab = 2 sqrt 3b\)
Therefore... \(\displaystyle a = (2 sqrt 3b)/2\)
Substitute it in:
\(\displaystyle ((2 sqrt 3b)/2)^2 - b^2 = 2\)
\(\displaystyle (4 sqrt 9)/4b^2 - b^2 = 2\)
\(\displaystyle 12/4b^2 -b^2 = 2\)
\(\displaystyle 12 - b^2 = 8b^2\)
\(\displaystyle 0 = 8b^2 + b^2 - 12\)
\(\displaystyle 0 = 9b^2 - 12\)
Ok Stuck
I'd be very happy if someone could help.
Oh also, in the same question, which I did a different way, I got \(\displaystyle a^2 + 2a = 3\) ---it's obvious that a is one, but how do I get from there to an answer?
Thanks alot
-Adam