Question

For an analytic function

\(f\left( {x + iy} \right) = \mu \left( {x,y} \right) + i\nu \left( {x,y} \right),\mu\) is given by

μ = 3x² – 3y²

Then expression for ν (considering K to be constant) is
1. 3y² – 3x² + k
2. 6x – 6y + k
3. 6x + 6y + k
4. 6xy + k

Answer 1

Correct Answer - Option 4 : 6xy + k

μ = 3x² – 3y²

\(\frac{{\partial \mu }}{{\partial x}} = 6x\ \&\ \frac{{\partial \mu }}{{\partial y}} = - 6y\) 

By Cauchy-Riemann equation

\(\begin{array}{l} \frac{{\partial \mu }}{{\partial x}} = \frac{{\partial \nu }}{{\partial y}}\ \&\ \frac{{\partial \mu }}{{\partial y}} = - \frac{{\partial \nu }}{{\partial x}}\\ \therefore \frac{{\partial \nu }}{{\partial y}} = 6x\ \&\ \frac{{\partial \nu }}{{\partial x}} = 6y \end{array}\) 

We know

\(dv = \frac{{\partial \nu }}{{\partial x}}dx + \frac{{\partial \nu }}{{\partial y}}dy\) 

⇒ dv = 6ydx + 6xdy

On integration

ν = 6xy + k      where k is constant