1. ## complex number help

I have a couple of questions that i cant figure out to save myself =/

1.these are expressing in the form a+ib $\displaystyle \frac{1}{(2+i)(z+1)} = 5-3i$

i got z= -1/14 + i/14

2. $\displaystyle 2z^2=6iz + 3$
didnt know how to do this.

3. $\displaystyle (z)(zcon)+2(z-zcon)=10 +2i$ (where zcon is the conjugate of z)

okay and theres:
"$\displaystyle 1-zi$ is the root of the equation
$\displaystyle z^2+(i+1)z+k=0$

find the other root and the value of k."

now i assume you use the $\displaystyle \Delta$ but im not sure.

im sorry theres so many but i need the help.
also if this isnt the correct forum also sorry as there was a similar problem listed here too

The first one is incorrect.

$\displaystyle \frac{1}{(2+i)(z+1)}=5-3i$

$\displaystyle 1=(5-3i)(2+i)(z+1))$

$\displaystyle 1=(13-i)(z+1)$

$\displaystyle z+1=\frac{1}{(13-i)}\times\frac{13+i}{13+i}$

Hope you will be able to continue.

The second one is a quadratic equation of z.

$\displaystyle 2z^2=6iz + 3\Rightarrow{2z^2-6iz - 3=0}$

$\displaystyle z^2-3iz-\frac{3}{2}=0$

$\displaystyle (z-\frac{3i}{2})^2+\frac{9}{4}-\frac{3}{2}=0$

$\displaystyle (z-\frac{3i}{2})^2+\frac{9}{4}-\frac{3}{2}=0$

$\displaystyle (z-\frac{3i}{2})^2+\frac{3}{4}=0$

Can you continue from here??

3. ok thanks i understand the first one

is the second
+/- $\displaystyle \frac{3i}{2}+\frac{9}{4}$?

i think i got my last posted one as well.
its a 2 not a z
do you use the fact that the sum of the roots would be $\displaystyle let x=second root x+(1-2i)=i+1 ?$
therfore figuring out k?

4. that didnt post as expected

should be
"letx = the second root

then the eqn starting with x

5. is the first one$\displaystyle z=\frac{-157}{170}+\frac{i}{170}$??

is the first one$\displaystyle z=\frac{-157}{170}+\frac{i}{170}$??
Correct.

ok thanks i understand the first one

is the second
+/-$\displaystyle \frac{3i}{2}+\frac{9}{4}$ ?

Incorrect.

$\displaystyle (z-\frac{3i}{2})^2=-\frac{3}{4}$

$\displaystyle (z-\frac{3i}{2})=\pm{\frac{i\sqrt{3}}{2}}$

$\displaystyle z=\frac{3i}{2}\pm{\frac{i\sqrt{3}}{2}}$

$\displaystyle z=(\frac{3}{2}\pm\frac{\sqrt{3}}{2})i$

3. $\displaystyle (z)(zcon)+2(z-zcon)=10 +2i$ (where zcon is the conjugate of z)

Use $\displaystyle z=x+iy$ and simplify. You would end up with two equations with x and y; hence you would be able to solve them simultaneously.

okay and theres:$\displaystyle 1-zi$" is the root of the equation $\displaystyle z^2+(i+1)z+k=0$ find the other root and the value of k."

You could use the sum and product of roots of a quadratic equation. That is if $\displaystyle \alpha~and~\beta$ are the roots of the quadratic, $\displaystyle ax^2+bx+c=0$; then, $\displaystyle \alpha+\beta=-\frac{b}{a}~and~\alpha\beta=\frac{c}{a}$

1.these are expressing in the form a+ib $\displaystyle \frac{1}{(2+i)(z+1)} = 5-3i$
i got z= -1/14 + i/14
There is a much easier way to do this one.

Recall that $\displaystyle \dfrac{1}{z} = \dfrac{{\overline z }}{{\left| z \right|^2 }}$

So $\displaystyle z + 1 = \dfrac{1}{{2 + i}}\dfrac{1}{{5 - 3i}} = \dfrac{{2 - i}}{5}\dfrac{{5 + 3i}}{{34}}$.

8. would my way of working out k in question 4 be right?

If the two roots of this equation are $\displaystyle \alpha~and~\beta$ then, $\displaystyle \alpha+(1-2i)=-(1+i)~and~\alpha(1-2i)=k$. From these two equations you could find k.