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Question

# An analytic function of a complex variable z=x+iy is expressed as f(z)=u(z,y)+iv(x,y), where i=√−1. if u(x,y)=2xy, then v(z,y) must be

A
x2+y2+ constnat
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B
x2y2+ constant
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C
x2+y2+ constant
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D
x2y2+ constant
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Solution

## The correct option is C −x2+y2+ constantMethod I: ∵ it is given that f(z)=u+iv is analytic & we have als given that u=2xy i.e., real part is given so we can use case I of MILNE THOMSON method i.e., Step 1: ∂u∂y=2y≈ϕ1(x,y) Step 2: ϕ1 (z,0)=0 Step 3: ∂u∂y=y≈ϕ2 (x,y) Step 4: ϕ2 (z,0)=2z Step 5: f(z)=∫[ϕ1(z,0)−iϕ2(z,0)]dz+c =∫(0−i(2z)dz)+c=−iz2+c =−i[x2−y2+2ixy]+c u+iv=u+2xy+i(y2−x2)+c So v=y2−x2+c Method II: ∵ u=2xy⇒ux=2y & uy=2x So, by C - R equation, vy=2y & vx=−2x Now, by Total Derivative Concept in v′ ∵V=V(x,y) So, dv=(∂v∂x)dx+(∂v∂y)dy ⇒v=∫−2xdx+∫2ydy+C =−x2+y2+C Method III: f(x)=u(x,y)+i v(x,y) u(x,y)=2xy⇒2xy⇒(∂v∂x)=2y Using Cauchy-Riemann condition ∂u∂x=∂v∂y v=y2+f(x) .....(i) Again, ∂u∂y=−∂v∂x 2x=−f′(x) .... (ii) ⇒f(x)−x2+c So using (i) v=y2−x2+c

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