Show that an analytic function can not have a constant modulus without reducing to a constant.
Algorithm of Basic Concepts
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Show that an analytic function can not have a constant modulus without reducing to a constant.
2 319 просмотров · 3 г. назад
Algorithm of Basic Concepts
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2 319 просмотров · 3 г. назад
Complex Analysis Theorem from Analytic function
Statement/Theorem /Prove that :-
Show that an analytic function with constant modulus is constant.
OR
Show that an analytic function cannot have a constant modulus without reducing to a constant.
Solution/ Proof:-
Let a function f(z)= u+iv
According to Cauchy-Riemann equations
(∂u/∂x)=(∂v/∂y) and (∂u/∂y)=-(∂v/∂x)
........(1)
The modulus of the function
f(z)=u+iv
⇒|f(z)|=constant = c (say)
⇒√(u²+v²)=c
⇒u²+v²=c² ........(2)
Now Partially differentiating Equation (2) w.r.t. x and y
2u(∂u/∂x)+2v(∂u/∂x)=0
⇒u(∂u/∂x)+v(∂u/∂x)=0
∵(∂v/∂x)=-(∂u/∂y)
⇒u(∂u/∂x)-v(∂u/∂y)=0 ........(3)
And
2u(∂u/∂y)+2v(∂u/∂y)=0
⇒u(∂u/∂y)+v(∂u/∂y)=0
∵(∂v/∂y)=(∂u/∂x)
⇒u(∂u/∂y)+v(∂u/∂x)=0 ........(4)
Multiplying v and u in question (3) and (4) respectively and subtracting Equation (4) from equation (3) we get
vu(∂u/∂x)-v²(∂u/∂x)-vu(∂u/∂y)-v²(∂u/∂x)=0
⇒-2v²(∂u/∂x)=0
⇒(∂u/∂x)=0
Putting in equation (3) we get
⇒(∂u/∂y)=0
And by( Cauchy-Riemann) equations (1)
⇒(∂u/∂x)=(∂v/∂y)=0 and ⇒-(∂u/∂y)=(∂v/∂x)=0
Here
(∂u/∂x),(∂u/∂y),(∂v/∂x)and(∂v/∂y)are equal to zero.
This shows that u and v are constant and function is f(z)=u+iv = constant.
Hence Proved...!!!
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