# Thread: Help with complex numbers

1. ## Help with complex numbers

Hi please see the attached pdf. I need help with the following

10.) B

11.)

12

13.

I'm confused. I already know all the answers because they are given. What i need is an explanation/guidance.
Thanks

2. ## Re: Help with complex numbers

10.)

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

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

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

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

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

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

Equating coefficients, we obtain the system:

$\displaystyle x+y=8$

$\displaystyle 3x-y=4$

$\displaystyle 4x=12$

$\displaystyle x=3$ and so

$\displaystyle y=5$

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

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

$\displaystyle z=2-2i$

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

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

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