No Idea How to do this...

[Iamadam]

I have the question:

Solve for the real numbers a and b...
(a+bi)^2 = 4 +3i

So what I did was....
(a+bi)(a+bi) = 4 + 3i

a^2 - b^2 + 2abi = 4 + 3i

(a+b)(a-b) +2abi = 4 + 3i

But I don't know where to go from here
(By the way the 'i' signifies an 'imaginary' number(i = sqrt(-1))

If someone could help I'd really appreciate it
Thanks,
-Adam

 

[galactus]

You have the answer staring at you:

\(\displaystyle a^{2}-b^{2}=4\)

\(\displaystyle 2ab=3\)

2 variables in 2 unknowns. Solve the system.

 

[Iamadam]

I also got up to that stage but wasn't sure if I'd done it right. How do you solve that?


\(\displaystyle a^2 - b^2 = 4\)

\(\displaystyle 2ab = 3\)

\(\displaystyle b = 3/2a\)

\(\displaystyle a^2 - (3/2a)^2 = 4\)

\(\displaystyle (a^2) - (9/4a^2) = 4\)

Is that all right? Where did I go wrong? And where do I go from there?
Ahh so many questions :shock:
I'm in a bit of a dumb mood and I just can't figure it out
-Adam

 

[galactus]

Just a little further.

You could multiply by \(\displaystyle 4a^{2}\) to get rid of the fraction and get:

\(\displaystyle 4a^{4}-16a^{2}-9\)

Now, you could factor, but the power of four is intimidating.

What you can do is a little subbing:

Let \(\displaystyle u=2a^{2}\)

You have: \(\displaystyle (2a^{2})^{2}-8(2a^{2})-9=0\)

=\(\displaystyle u^{2}-8u-9=0\)

Factor: \(\displaystyle (u-9)(u+1)=0\)

Sub back in: \(\displaystyle (2a^{2}-9)(2a^{2}+1)=0\)

Solve \(\displaystyle 2a^{2}-9=0\) and \(\displaystyle 2a^{2}+1=0\)

Use the appropriate subbing to find b.