Complex Variables and Partial Differential Equations (3130005)

Question-3b

BE | Semester-3 Winter-2019 | 26-11-2019

Q3) (b)

For,

f (z) = 3 x + y + i (3 y - x)

\Rightarrow u = 3 x + y & v = 3 y - x

\Rightarrow \frac{\partial u}{\partial x} = 3; \frac{\partial u}{\partial y} = 1; \frac{\partial v}{\partial x} = - 1; \frac{\partial v}{\partial y} = 3

\Rightarrow \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} & \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x} .

⇒ CR - equations are satisfied.

\Rightarrow f (z)

is analytic function.

For,

f (z) = z^{\frac{3}{2}}

For,

f (z) = {[r (\cos θ + i \sin θ)]}^{\frac{3}{2}}

For,

f (z) = r^{\frac{3}{2}} {(\cos θ + i \sin θ)}^{\frac{3}{2}}

For,

f (z) = r^{\frac{3}{2}} (\cos \frac{3 θ}{2} + i \sin \frac{3 θ}{2})

For,

f (z) = r^{\frac{3}{2}} \cos \frac{3 θ}{2} + i r^{\frac{3}{2}} \sin \frac{3 θ}{2}

\Rightarrow u = r^{\frac{3}{2}} \cos \frac{3 θ}{2} & v = r^{\frac{3}{2}} \sin \frac{3 θ}{2}

\Rightarrow \frac{\partial u}{\partial r} = \frac{3}{2} r^{\frac{1}{2}} \cos \frac{3 θ}{2} & \frac{\partial u}{\partial θ} = - \frac{3}{2} r^{\frac{3}{2}} \sin \frac{3 θ}{2}

\Rightarrow \frac{\partial v}{\partial r} = \frac{3}{2} r^{\frac{1}{2}} \sin \frac{3 θ}{2} & \frac{\partial v}{\partial θ} = \frac{3}{2} r^{\frac{3}{2}} \cos \frac{3 θ}{2}

\Rightarrow \frac{\partial u}{\partial r} = \frac{1}{r} \frac{\partial v}{\partial θ} & \frac{\partial u}{\partial θ} = - r \frac{\partial v}{\partial r}

⇒ CR - equations are satisfied.

\Rightarrow f (z)

is analytic function.