Question

An analytic function of a complex variable $z=x+iy$ is expressed as $f\left(z\right)=u(z,y)+iv(x,y),$ where $i=\sqrt{-1}$. if $u(x,y)=2xy,$ then $v(z,y)$ must be

• A. $x^2+y^2+$ constnat
• B. $x^2-y^2+$ constant
• C. $-x^2+y^2+$ constant
• D. $-x^2-y^2+$ constant

Answer 1

The correct option is C $-x^2+y^2+$ constant

Method I:
$\because$ it is given that $f\left(z\right)=u+iv$ is analytic & we have als given that $u=2xy$ i.e., real part is given so we can use case I of MILNE THOMSON method i.e.,

Step 1: $\frac{\partial u}{\partial y}=2y\approx {\varphi }_1(x,y)$
Step 2: ${\varphi }_1(z,0)=0$
Step 3: $\frac{\partial u}{\partial y}=y\approx {\varphi }_2(x,y)$
Step 4: ${\varphi }_2(z,0)=2z$
Step 5: $f\left(z\right)=\int \left[{\varphi }_1\right(z,0)-i{\varphi }_2(z,0\left)\right]dz+c$
$=\int (0-i(2z\left)dz\right)+c=-i{z}^2+c$
$=-i[x^2-y^2+2ixy]+c$
$u+iv=u+2xy+i(y^2-x^2)+c$
So $v=y^2-x^2+c$
Method II:
$\because$ $u=2xy\Rightarrow u_x=2y$ & $u_y=2x$
So, by C - R equation,
$v_y=2y$ & $v_x=-2x$
Now, by Total Derivative Concept in $v^{\prime }$
$\because V=V(x,y)$
So, $dv=\left(\frac{\partial v}{\partial x}\right)dx+\left(\frac{\partial v}{\partial y}\right)dy$
$\Rightarrow v=\int -2xdx+\int 2ydy+C$
$=-x^2+y^2+C$
Method III:
$f\left(x\right)=u(x,y)+iv(x,y)$
$u(x,y)=2xy\Rightarrow 2xy\Rightarrow \left(\frac{\partial v}{\partial x}\right)=2y$
Using Cauchy-Riemann condition
$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$
$v=y^2+f\left(x\right)$ .....(i)
Again, $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$
$2x=-f^{\prime }\left(x\right)$ .... (ii)
$\Rightarrow f\left(x\right)-x^2+c$
So using (i) $v=y^2-x^2+c$