Question

Suppose z = x + iy = r*(e^(-i*theta)) , where i=sqrt(=1) , r =|z| , tan(theta) = y/x

Also, suppose that w = -uz - m*ln(z) = A + Bi where i=sqrt(-1) and ln=log_e

Then |dw/dz| = 0 implies, -(u + (m/z) =0 ==> z =-(m/u)

Since w = -uz - m*ln(z) = A + Bi ,

A + Bi = -uz - mln(z) = -(ur(e^(itheta))) - mln(r*(e^(i*theta)))

==> A+ Bi = -uz -m[ln(r) - ln(e^(-itheta)) ]
==> A+ Bi = -uz -mln(r) + mln(e^(-itheta))
==> A + Bi = -uz - mln(r) -(mtheta)i

Let theta = T

Then,
A + Bi = -ur(cos(T) - isin(T)) -mln(r) - (mT)i

which implies,
B = ursin(T) -mT
==> B = uy - marctan(y/x) , since r*sin(theta) = y and tan(theta) =y/x

But, the solution is given as B = -uy - m*arctan(y/x)

Could anyone point out the mistake in my calculation?

Thank you!

Answer 1

Your first line bothers me. You say
z = x + iy = r*(e^(-iθ))
But
z = x + iy = r*(e^(iθ))

Answer 2

My professor has given so in the note, I guess that might be the reason for this confusion.
Thank you for pointing that out!