hi, could someone check the answers of these for me? btw, the book says that they are all wrong. i dont understand why.

1. If z = 2 + i, evaluate:

a) 3z = 6 + 3i

b) $\displaystyle (z-1)(z^2 + z +1) $

$\displaystyle = z^3 - 1 $

$\displaystyle = (2+i)^3 $

$\displaystyle = (2 + i)^2(2+i) $

$\displaystyle = (4 + i^2 + 4i)(2+2i) $

$\displaystyle = (3 + 4i) (2+2i) $

$\displaystyle = 6 + 8i + 6i + 8i^2 $

$\displaystyle = -2 + 14i$

2. Solve for z; expressing answers in the form a + ib

a) $\displaystyle \frac{z+3}{z - 1} = 2 - 3i $

$\displaystyle \frac{z -1 + 4}{z-1} = 2-3i$

$\displaystyle 1 + \frac{4}{z-1} = 2 - 3i$

$\displaystyle 2 - 3i -1 = \frac{4}{z-1}$

$\displaystyle 1 - 3i = \frac{4}{z -1} $

$\displaystyle \frac{4}{1-3i} + 1 = z $

$\displaystyle \frac{4(1 + 3i)}{(1-3i)(1+3i)}+ 1 =z$

$\displaystyle \frac{4 + 12i}{10} + 1 = z$

$\displaystyle \frac{4+12i + 10}{10} = z $

$\displaystyle \frac{14 + 12i}{10} = z $

$\displaystyle \frac{7}{5} + \frac{6}{5}i = z$

b) Totally clueless about this one:

Solve for x. $\displaystyle \frac{2z}{2+i} + 3- 2i = (1-i)z $

3. Simplify: $\displaystyle \frac{\sqrt{5+12i}+ \sqrt{5-12i}}{\sqrt{5+12i}-\sqrt{5-12i}} $

I got 3i/2. but the answer is -3i/2. I dont get why.

Thanks