complex number help

**link0099** · October 27th 2010, 11:16 PM

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

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

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

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

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

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

find the other root and the value of k."

now i assume you use the $\Delta$ 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

**Sudharaka** · October 28th 2010, 12:35 AM

Dear link0099,

The first one is incorrect.

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

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

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

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

Hope you will be able to continue.

The second one is a quadratic equation of z.

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

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

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

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

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

Can you continue from here??

**link0099** · October 28th 2010, 01:28 AM

ok thanks i understand the first one

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

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 $let x=second root<br /> x+(1-2i)=i+1 ?$
therfore figuring out k?

**link0099** · October 28th 2010, 01:32 AM

that didnt post as expected

should be
"letx = the second root

then the eqn starting with x

**link0099** · October 28th 2010, 02:17 AM

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

**Sudharaka** · October 28th 2010, 05:03 AM

Originally Posted by link0099

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

Correct.

ok thanks i understand the first one

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

Incorrect.

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

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

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

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

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

Use $z=x+iy$ 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: $1-zi$ " is the root of the equation $z^2+(i+1)z+k=0$ 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 $\alpha~and~\beta$ are the roots of the quadratic, $ax^2+bx+c=0$ ; then, $\alpha+\beta=-\frac{b}{a}~and~\alpha\beta=\frac{c}{a}$

**Plato** · October 28th 2010, 06:26 AM

Originally Posted by link0099

1.these are expressing in the form a+ib $\frac{1}{(2+i)(z+1)} = 5-3i$
i got z= -1/14 + i/14

There is a much easier way to do this one.

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

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

**link0099** · October 28th 2010, 08:40 PM

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

**Sudharaka** · October 29th 2010, 02:40 AM

Originally Posted by link0099

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

Dear link0099,

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

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