# GATE2017 ME-2: 29

If $f(z)=(x^{2}+ay^{2})+i bxy$ is a complex analytic function of $z=x+iy$, where $i=\sqrt{-1}$, then

1. $a=-1, b=-1$
2. $a=-1, b=2$
3. $a=1, b= 2$
4. $a=2, b=2$
in Calculus
recategorized

Let, $f(z) = f(x,y) = u(x,y) + iv(x,y)$ be a complex function.
Now, In this question, $u(x,y)= x^2 + ay^2$ and $v(x,y)= bxy$

So, $\frac{\partial u}{\partial x}=2x, \; \frac{\partial^2 u}{\partial x^2}=2, \; \frac{\partial u}{\partial y}=2ay, \;\frac{\partial^2 u}{\partial y^2} =2a\;$ and

$\frac{\partial v}{\partial x}=by, \; \frac{\partial^2 v}{\partial x^2}=0, \; \frac{\partial u}{\partial y}=bx, \;\frac{\partial^2 u}{\partial y^2} =0\;$

Now, If $f(z)$ is analytic then $\frac{\partial^2 u}{\partial x^2}$ $+$ $\frac{\partial^2 u}{\partial y^2} = 0$ and $\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}$

So, $2+2a=0$ and $2x=bx$

$\Rightarrow$ $a=-1, \; b= 2$

So, Answer should be $(B)$

410 points 1 2 4

## Related questions

The surface integral $\int \int _{s} F.n$ dS over the surface $S$ of the sphere $x^{2}+y^{2}+z^{2}=9$, where $F=(x+y) i+(x+z) j+(y+z)k$ and $n$ is the unit outward surface normal, yields ________.
The divergence of the vector $-yi+xj$ is ________.
Value of $\left ( 1+i \right )^{8}$, where $i=\sqrt{-1}$, is equal to $4$ $16$ $4i$ $16i$
The function $f(z)$ of complex variable $z=x+iy$, where $i=\sqrt{-1}$, is given as $f(z)=(x^3-3xy^2)+i \: v(x,y)$. For this function to be analytic, $v(x,y)$ should be $(3xy^2-y^3) +$ constant $(3x^2y^2-y^3) +$ constant $(x^3-3x^2 y) +$ constant $(3x^2y-y^3) +$ constant
An analytic function $f(z)$ of complex variable $z=x+iy$ may be written as $f(z)=u(x,y)+iv(x,y)$. Then $u(x,y)$ and $v(x,y)$ ...