Question

Q.Which of following function is analytic in comolex plane.
(i) f(z)=z ̅ (ii) f(z)=2x+ixy^2 (iii) f(z)= z^2

Answer 1

A complex-valued function can be represented as $f(z)=u(x,y)+iv(x,y)$ . $f(z)$ is an analytic function in complex plane, if $u(x,y)$ and $v(x,y)$ satisfy Cauchy-Riemann equations at any point, i.e. $\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}$ , $\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$ .

(i) $f(z)=\overline{z}=x-iy$

$\frac{\partial u}{\partial x}=1$, $\frac{\partial v}{\partial y}=-1$. The two partial derivatives are not equal. So $f(z)=\overline{z}$ is not analytic.

(ii) $f(z)=2x+ixy^2$

$\frac{\partial u}{\partial x}=2$, $\frac{\partial v}{\partial y}=2xy$. The two partial derivatives are not equal. So $f(z)=2x+ixy^2$ is not analytic.

(iii) $f(z)=z^2=x^2-y^2+2ixy$

$\frac{\partial u}{\partial x}=2x, \frac{\partial v}{\partial y}=2x, \frac{\partial u}{\partial y}=-2y, \frac{\partial v}{\partial x}=2y.$

So $u(x,y)$ and $v(x,y)$ satisfy Cauchy-Riemann equations at any point of complex plane, hence $f(z)=z^2$ is analytic.

Answer: (iii) $f(z)=z^2$