Question here?

Nothing too difficult now....:-)

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

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

infinitely many solutions

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

hi

I would write z=a+bi and w=c+di expand ...equate real and imaginary parts...then solve the system

yes both are complex

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

o.O

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

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

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

no a+bi

(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

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

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