Question

The real part of an analytic funciton $f\left(z\right)$ where $z=x+jy$ is givne by $e^{-y}cos\left(x\right)$. THe imaginary part of $f\left(x\right)$ is

• A. $e^{y}cos\left(x\right)$
• B. $e^{-y}sin\left(x\right)$
• C. $-e^{y}sin\left(x\right)$
• D. $-e^{-y}sin\left(x\right)$

Answer 1

The correct option is B $e^{-y}sin\left(x\right)$

Methdo I: Using MILNE THOMSON method
$f\left(z\right)=u+iv$ (Analytic)
Where $u=e^{--y}cos\left(x\right)$ (Real part)
Step 1: $\frac{\partial u}{\partial x}=-e^{y}sin\left(x\right)\approx {\varphi }_1(x,y)$
Step 2: ${\varphi }_1(x,0)=-sinz$
Step 3: $\frac{\partial u}{\partial y}=-e^{x}cosx\approx {\varphi }_2(x,y)$
Step 4: ${\varphi }_2(z,0)=-cosz$
Step 5: $f\left(z\right)=\int \left[{\varphi }_1\right(z,0)-i{\varphi }_2(z,0\left)\right]dz+c$
$=\int (-sinz+icosz)dx+c$
$=(cosz+isinz)+c$
$=e^{iz}+c$ {$\because e^{i\theta }=cos\theta +isin\theta$}
$=e^{i(x+iy)}+c$
$=e^{ix-y}+c=e^{-y}\left(e^{ix}\right)+c$
$u+iv=e^{-y}(cosx+isnx)+c$
Hence, $v=e^{-y}sinx$ (for any $c=0$)
Method II:
Using total derivative concept
$u=e^{-y}cosx$ then $\frac{\partial u}{\partial x}=-e^{-y}sinx$ and $\frac{\partial u}{\partial y}=-e^{-y}cosx$
$\because v=v(x,y)$
$dv=\left(\frac{\partial v}{\partial x}\right)dx+\left(\frac{\partial v}{\partial y}\right)dy$
$=(-\frac{\partial u}{\partial y})dx+\left(\frac{\partial u}{\partial x}\right)dy$ (By C-R equation)
$=(+e^{-y}cosx)dx+(-e^{-y}sinx)dy$
$dv=d\left(e^{-y}sinx\right)$
On integraing, $v=e^{-y}sinx+c$
If we take$c=0$ then $v=e^{-y}sinx$
Method III:
Let $f\left(z\right)=u+iv$
$u=e^{-y}cosx$
$\Rightarrow u_x=-e^{-y}sinx$ & $u_y=-e^{-y}cosx$
By C.R. eqns. $u_x=v_y$
$v_y=-e^{-y}sinx$
Interating both sides w.r.t $y$
$v=e^{-y}sinx+\varphi \left(x\right)$
By C.R.eqn.'s
$u_y=-v_y$
$\Rightarrow -e^{-y}cosx=-e^{-y}cos-{\varphi }^{\prime }\left(x\right)$
${\varphi }^{\prime }\left(x\right)=0$
$\varphi \left(x\right)=C$
so. $v=e^{-y}sinx+C$
Note: We can also use MILNE THOMSON method to find $v$.