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.
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?