Find all complex numbers z such that z^2=15−8i, and give your answer in the form a+bi.
I'm not sure if I have this right

Question

Find all complex numbers z such that z^2=15−8i, and give your answer in the form a+bi.
I'm not sure if I have this right

anonymous · Answer

$$a^2-b^2+2abi=15 -8i$$
$$a^2-b^2=15, 2ab=-8$$
$$2ab=-8, b=-4a$$
$$a^2-(16a^2)=15$$
$$-15a^2=15, a^2=-1$$
$$a= \pm i$$
$$b=-4(i)=-4i$$
$$b=-4(-i)=4i$$
roots:
z= -i+4i=3i
z=i-4i=-3i
is this right cuz it does not seem right

anonymous · Answer

@ganeshie8

anonymous · Answer

@freckles

freckles · Answer

I got to the third line and seen something just a bit off.

freckles · Answer

2ab=-8 
solving this for b gives:
b=-4/a   not b=-4a

anonymous · Answer

ooh your right, no wonder I was getting something way off

anonymous · Answer

oh ok now it makes sense, now it gives me a quartic function and now I can solve for a to get b

freckles · Answer

yeah did you get:
$$a^4-15a^2-16=0 \ (a^2-16)(a^2+1)=0 ?$$

anonymous · Answer

yea

freckles · Answer

cool you will actually get some real values for a which is what you were actually looking for

anonymous · Answer

Yes thank you :)