Solving Complex Math: 2 Questions Answered

[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.

 

[radou]

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.