1. ## Complex numbers

simplify 1+2j/1+j

I got 1+2j/1+j=(1+2j)(1-j)/(1+j)(1-j)=3+j/2
is this right?

and
z^2+4z+13=0
i have no idea how to do this.... can someone tell me how to start?
thanks.

2. Originally Posted by MK47
simplify 1+2j/1+j

I got 1+2j/1+j=(1+2j)(1-j)/(1+j)(1-j)=3+j/2
is this right?
It is!

Originally Posted by MK47
and
z^2+4z+13=0
i have no idea how to do this.... can someone tell me how to start?
thanks.

Have you tried the quadratic formula?

3. Have you tried the quadratic formula?[/QUOTE]
i got -2-3j and -2+3j

another question that i am kind of stuck on
show that 3+2j is a root of the quation z^3+7z^2+19z-13=0
and find the other two roots.

this is what i thought i should do but it's wrong....
z=(3+2j)
z^2=(3+2j)(3+2j)
=5+12j
z^3=(5+12j)(3+2j)
=-9+46j

(-9+46j)+7(5+12j)+19(3+2j)-13=0
66+148j=0
but it doesn't......
i know the other root would be 3-2j but what did i do wrong?

4. You aren't doing anything wrong unless you copied the problem wrong. 3+ 2j is NOT a root of $z^3+7z^2+19z-13=0$.

5. well i just checked and it is 3+2j and i put the equation into my calculator ,it gave me 0.56,-3.78+2.99i and -3.78-2.99j..... so i guess my teacher wrote the question wrong.