gentle silo
So context, for part (a), it is differentiable only at (0,0) and thus is not analytic anywhere
Now my issue is part (b), let a = k + ij, where k and j are real.
f(z) = k(x^2 + y^2) + v(x,y) + ij(x^2 + y^2)
Suppose f(z) is analytic, then k(x^2 + y^2) + v(x,y) and j(x^2 + y^2) are harmonic conjugates
since v is not specified, let's try looking at j(x^2 + y^2).
1st partial wrt x of j =2jx; double partial wrt x of j = 2j
1st partial wrt y of j =2jy; double partial wrt y of j = 2j
If j is harmonic, then the sum of the 2nd derivatives = 0. i.e 2j + 2j = 0 => j = 0
but that means f(z) is real only, since Im(fz) = ij(x^2 + y^2)
if f(z) is always real, then i do not think it can be analytic because solution of CR equations for f(z) would be 0
I am at wits end. Help!
gray acorn