Help with complex numbers

10.)

b) $\frac{1}{2-i}-\frac{1}{2+i}=\frac{(2+i)-(2-i)}{2^2-i^2}=\frac{2i}{5}=0+\frac{2}{5}i$

11.) $\frac{x}{1-i}+\frac{y}{1+3i}=2$

$\frac{x(1+3i)+y(1-i)}{(1-i)(1+3i)}=2$

$\frac{x+3xi+y-yi}{1+2i-3i^2}=2$

$\frac{(x+y)+(3x-y)i}{4+2i}=2$

$(x+y)+(3x-y)i=8+4i$

Equating coefficients, we obtain the system:

$x+y=8$

$3x-y=4$

Adding, we find:

$4x=12$

$x=3$ and so

$y=5$

12.) $(z+i)(2-i)=3-4i$

$z+i=\frac{3-4i}{2-i}\cdot\frac{2+i}{2+i}=\frac{10-5i}{5}=2-i$

$z=2-2i$

13.) $k(x-(2+3i))(x-(2-3i))=k((x-2)-3i)((x-2)+3i)=$

$k\left((x-2)^2-(3i)^2 \right)=k\left(x^2-4x+4+9 \right)=k\left(x^2-4x+13 \right)=0$

where $0\ne k\in\mathbb{R}$