Question

how do i factor Z^2+3i

Answer 1

over what? complex numbers?

Answer 2

yes

Answer 3

lorda mercy you have to find \(\sqrt{3i}\) then right?

Answer 4

yes lol exactly

Answer 5

doesn't i^2 result in a negative number though?

Answer 6

yes i^2 =-1...but I have to factor 3i, not square it

Answer 7

well by the looks of things you have to find the square root of negative 3i

Answer 8

to find z

Answer 9

actually even that is not right is it? if it was
\[z^2-3i\] then you would factor as \(z+\sqrt{3i})(z-\sqrt{3i})\)

Answer 10

you actually need \(\sqrt{-3i}\)

Answer 11

which is not really any different from finding \(\sqrt{3i}\)

Answer 12

your right sat but 3i is positive.....(z^2+3i)

Answer 13

lol no \(3i\) is not positive

Answer 14

write in trig form or exponential form as
\[3\cos(\frac{3\pi}{2}+i\sin(\frac{3\pi}{2}))\] then take the real square root of 3 and divide the angle by 2

Answer 15

one of the answers will be
\[\sqrt{3}\left (\cos(\frac{3\pi}{4})+i\sin(\frac{3\pi}{4})\right)\]

Answer 16

the other will be half way across the circle

Answer 17

you good from there?

Answer 18

I think I follow, but it somewhat leads me back to my original statement about the factors being positive

Answer 19

complex numbers are not positive or negative

Answer 20

suppose you wanted to factor \(z^2+4\)
you could do this by writing
\[z^2+4=0,z^2=-4,z=\pm\sqrt{-4}=\pm2i\] and so
\[z^2+4=(z+2i)(z-2i)\]

Answer 21

because \(-4\) has two complex roots, \(2i\) and \(-2i\)
you have to do the same thing here, except instead of 4 you have \(3i\)

Answer 22

meaning you want the two complex roots of \(-3i\)

Answer 23

got you

Answer 24

ok

Answer 25

yes and thank you so much.

Answer 26

yw