Mumbai university > Electronics and telecommunication engineering, Electronics engineering > Sem 3 > Applied mathematics 3

Marks : 06

Years : MAY 2016

$$f(z) =u +iv \Rightarrow f'(z) =\dfrac {du}{dx} + i\dfrac {dv}{dx}=\dfrac {du}{dx} -i\dfrac {du}{dy} \text{( By C.Real )} $$

$\dfrac {du}{dx}=\dfrac 12 \log (x^2+y^2) + \dfrac x2 \dfrac 1{x^2+y^2} (2x) -y \dfrac 1{1+\dfrac {y^2}{x^2}}[-\dfrac y{x^2} ]+\cos x \cos hy \\ (\dfrac {du}{dx})_{x=z,y=0} =\dfrac 12 \log (z^2) +\dfrac {z^2}{z^2} + 0 + \cos z =\log z + 1 + \cos z \\ \dfrac {du}{dy}=\dfrac x2 \dfrac {(2y)}{(x^2+y^2)}-\tan^{-1} (y/x) -y \dfrac 1{1+y^2/x^2}[-\dfrac y{x^2}] + \sin x \sin hy$

by Milne Thompson method

$(\dfrac {du}{dy})_{x=z,y=0} = 0-0+0+0=0 \\ f'(z) =\log z + 1 + \cos z -i(0) \\ f(z) =\int dz + \int \log z \space dz + \int \cos z \space dz =z + z.\log z -\int dz + \sin z \\ f(z) =z\log z + \sin z + c$