
Bradford T. answered 11/27/20
Retired Engineer / Upper level math instructor
-uz-ln(z) = -uz-ln(re-iθ) = -uz - (ln(r)+ln(e-iθ) = -uz - (ln(r)-iθ) = -uz -ln(r) + iθ not -uz-ln(r) -iθ
Ashley P.
asked 11/27/20Suppose 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!
Bradford T. answered 11/27/20
Retired Engineer / Upper level math instructor
-uz-ln(z) = -uz-ln(re-iθ) = -uz - (ln(r)+ln(e-iθ) = -uz - (ln(r)-iθ) = -uz -ln(r) + iθ not -uz-ln(r) -iθ
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