An analytic function of a complex variable z = x + iy is expresse
A. xy + c
B. <span class="math-tex">\(\frac{{{x^2} + {y^2}}}{2} + c\)</span>
C. 2xy + c
D. <span class="math-tex">\(\frac{{{{\left( {x - y} \right)}^2}}}{2} + c\)</span>
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Right Answer is: C
SOLUTION
Concept:
f(z) = u(x, y) + i v(x, y)
since f(z) is analytic, C-R equations are as follows
\(\frac{{\partial u}}{{\partial x}} = \frac{{\partial v}}{{\partial y}}\;and~\frac{{\partial u}}{{\partial y}} = - \frac{{\partial v}}{{\partial x}}\)
Calculation:
Given:
u = x2 – y2
\(\frac{{\partial u}}{{\partial x}} = 2x = \frac{{\partial v}}{{\partial y}}\)
\(\frac{{\partial u}}{{\partial y}} = - 2y = - \frac{{\partial v}}{{\partial x}}\)
\(\therefore dv = \frac{{\partial v}}{{\partial x}}dx + \frac{{\partial v}}{{\partial y}}dy\)
⇒ dv = 2y dx + 2x dy
⇒ v = 2xy + c
Alternate method:
\({\rm{u}} = {{\rm{x}}^2} - {{\rm{y}}^2}\)
\(\frac{{{\rm{du}}}}{{{\rm{dx}}}} = 2{\rm{x~~and}}~~\frac{{{\rm{du}}}}{{{\rm{dy}}}} = {\rm{}} - 2{\rm{y}}\)
Since f(z) is analytic, C-R equations are as follows
\(\frac{{{\rm{du}}}}{{{\rm{dx}}}} = \frac{{{\rm{dv}}}}{{{\rm{dy}}}}{\rm~~{and}}~~\frac{{{\rm{du}}}}{{{\rm{dy}}}} = - \frac{{{\rm{dv}}}}{{{\rm{dx}}}}\)
We know that, f’(z) = \(\frac{{{\rm{du}}}}{{{\rm{dx}}}} + {\rm{i}}\frac{{{\rm{dv}}}}{{{\rm{dx}}}} = \frac{{{\rm{du}}}}{{{\rm{dx}}}} + {\rm{i}}\frac{{ - {\rm{du}}}}{{{\rm{dy}}}} = 2{\rm{x}} + {\rm{i}}\left( {2{\rm{y}}} \right)\)
Put x = z and y = 0
f’(z) = 2z
Integrate both side,
\(\mathop \smallint \nolimits {\rm{f'}}\left( {\rm{z}} \right) = {\rm{}}\mathop \smallint \nolimits 2{\rm{zdz}}\)
f(z) = z2 + c
For imaginary part,
f(u + iv) = (x + iy)2 + c = (x2 – y2 + i(2xy)) + c
Therefore, u = (x2 – y2) and v = 2xy