hey guys, i have some difficulties when doing FP1, which says
Find the square roots of 5-12i
The following is my working
finally i get the answer but i have one question
i. why the final answer is not ±(3 +2i)?
please someone help!
by the way, the answer is ±(3 − 2i)
Last edited by 041087; 01-07-2012 at 08:04.
You're not far off the full solution, though you made a bit of a mess when solving the quartic.
Firstly, a,b are real.
So when you had your solutions for b, you can discard the two complex ones, and are just left with b=2 or -2.
Then using a= -6/b, you can work out the "a" value in each case, and you then have the two desired solutions. Done.
When solving the quartic, you should recognize that this is a quadratic in b^2. You could use a substitution of c=b^2, though it's not necessary.
So factorising you get (b^2-4)(b^2+9) = 0
Since b is real, b^2 is positive, and so the only solution is b^2=4, and hence b=+/- 2....
Edit: In response to your last point.
If b=2, then a=-3, so one solution is -3+2i
If b=-2, then a=3, so other solution is 3-2i
Hence +/- (3-2i)
Last edited by ghostwalker; 01-07-2012 at 08:18.