Question

Question here?

Answer 1

Nothing too difficult now....:-)

Answer 2

\[If z,w \in C so that z ^{2}+w^{2}-2i(z-w)=2\] find z,w

Answer 3

\[z^2+w^2-2i(z-w)=2\] and i take it that w and z are complex right?

Answer 4

infinitely many solutions

Answer 5

i think i messed this up once before! hello zarkon!

Answer 6

hi

Answer 7

I would write z=a+bi and w=c+di

expand ...equate real and imaginary parts...then solve the system

Answer 8

yes both are complex

Answer 9

It's a complex difference of squares, myininaya can solve it..:-)

Answer 10

o.O

Answer 11

what a lot of algebra. i bet i messed up already

Answer 12

its easy... do u want me to solve it?

Answer 13

is there a snap way or do i have to slog through the
\[a+bi\] stuff?

Answer 14

no a+bi

Answer 15

(z-i)2+(w+i)2=0, (z-i)2-i2(w+i)2=0, (z-i)2-(iw-1)2=0, z=i+iw-1 or z=i-iw+1, with w ε C

Answer 16

\[(z-i)2+(w+i)2=0, (z-i)2-i2(w+i)2=0,
\]
\[(z-i)2-(iw-1)2=0,\]
\[ z=i+iw-1 or z=i-iw+1, with w ε C\]