# Complex Numbers

1. Oct 2, 2006

### suspenc3

2 Questions:

(1.)Carry out the indicated calculation:

$$(\frac{-6+2i}{1-8i})^2$$

=$$\frac{36-24i+4i^2}{1-16i+64i^2}$$

since $$i^2=-1$$

=$$\frac{32-24i}{-63-16i}[\frac{-63+16i}{-63+16i}]$$

I carry out the math and get an answer of:

$$\frac{-2400+1512i}{4225}$$

I must be doing something wrong with the conjugate, as far as i can tell it looks right, but for some reason my signs should be switched to give me the correct answer of:

$$\frac{-1632+2024i}{4225}$$

(2.)determine the set of all z satisfying the given equation or inequality:

$$|z-2i|<=|z+1+i| & |z|> 4$$

I solved this one down to:
$$x^2+y^2-4y+4<=x^2+y^2+2x+2y+2$$ & $$x^2+y^2>4$$

what do I do to simplify it?

Thanks.

2. Oct 2, 2006

Well, you just have to find the intersection of $$y \geq -\frac{1}{3}x+\frac{1}{3}$$ (implied by the first inequality) and $$x^2+y^2 > 4$$, i.e. $$S=\left\{(x,y) \in \textbf{R}^2: y \geq -\frac{1}{3}x+\frac{1}{3} \wedge x^2+y^2 > 4 \right\}$$. Draw a sketch of S.