A Radical Equation | Problem 462

[@scottleung9587]

Nice!

[@mcwulf25]

I thought you were going to substitute w = sqrt(2z) z = (w^2)/2 ((w^2)/2 + w)/2 = -1+i w^2 + 2w + 4(1-i) = 0 w = (-2 +- sqrt(4 - 16(1-i))) / 2 w = -1 +- sqrt(4i - 3) w = -1 +- sqrt((1+2i)^2) w = -1 +- (1+2i) w = 2i or w = -2(1+i) z = (2i)^2/2 = -2 or z = 4(1+i)^2 /2 = 4i Then it depends on what the square root sign means in the complex world.

[@shartokfancader8750]

|z-i|e^iarg(conj(z)+i)=z*(1-i)/2 Solve this one please

[@RuleofThehyperbolic]

i wonder if something like this could be solved? z^w = w^z where w is a known value