Question

solve z^2+(1-i)z+(-6+2i)

Answer 1

z^2+z-iz-6+26

Answer 2

find the value of z in terms of i

Answer 3

it is not an equation, there is nothing to 'solve'

Answer 4

if it was for example
\[z^2+(1-i)z+(-6+2i)=0\]then you could solve for \(z\) via the quadratic formula

Answer 5

oh its =0 i forgot to write it! im trying to solve it by Q.F but not getting my ans

Answer 6

by some miracle this one actually factors, as
\[(z-2)(z+(3-i))=0\] but that is not so easy to see

Answer 7

\[z=-\frac{(-1+i)+\pm\sqrt{(1+i)^2-4(1)(-6+2i)^2}}{2}\]

Answer 8

all the work is finding
\[(1-i)^2-4(-6+i)^2\]

Answer 9

\[(1-i)^2=1-2i-1=-2i\]

Answer 10

i dont get the underoot part

Answer 11

you have to compute
\[(1-i)^2-4(-6+2i)\] that is the hard work

Answer 12

also why have you changed the signs?

Answer 13

you get
\[1-2i-1+24-8i\]
\[=-10i

Answer 14

24-10i

Answer 15

changed the signs in front in the numerator because it is
\[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] and \(b=1-i\) so \(-b=-1+i\)

Answer 16

ok then what?

Answer 17

u there?

Answer 18

hmm i am thinking
your job now is to take the square root of \(24-10i\)

Answer 19

the answer is \(5-i\) but i am not sure how you are supposed to get it

Answer 20

thats the main part on which im stuck since the last 2 hours

Answer 21

i used wolfram
the only other way i know how to do this is to write the number in polar form, then to it

Answer 22

alright

Answer 23

or use a calculator that has complex numbers (most do now)

Answer 24

mine doesnt