Mumbai University > Information Technology, Computer Engineering > Sem 3 > Applied Mathematics 3

Marks: 6M

Year: Dec 2013

Since u is harmonic

$\frac{∂^2 u}{∂x^2} + \frac{∂^2 u}{∂y^2} = 0$

By data $f(z) = u_x – i u_y$

Let the $u_x = U$ and $– u_y = V$ , so that f(z) = U + iV.

We have to show that f(z) is analytic.

Now, $U_x = \frac{∂^2 u}{∂x^2} = - \frac{∂^2 u}{∂y^2}$ [By 1]

And $U_y = \frac{∂^2 u}{∂x∂y}$

$V_x = \frac{∂^2 u}{∂y∂x} \hspace{0.2cm} and \hspace{0.2cm} V_y = - \frac{∂^2 u}{∂y^2}$

$:. U_x = V_y \hspace{0.2cm} and \hspace{0.2cm} U_y = -V_x .$

$:. f(z) = U + i V \text{is analytic i.e.} f(z) = u_x – i u_y \text{is analytic.}$