[MCQ/CV-02001] Analytic functions
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If \(f(z)\) is an analytic function of \(z\), \(\left( \frac{\partial^2 }{\partial x^2} + \frac{\partial^2 }{\partial y^2}\right) \left| f(z)\right|^2\) is equal to
| (A) \(2 \big| f^ \prime(z)\big|^2\) |
(B) \(3 \big|f^\prime(z)\big|^2\) |
| (C) \(4 \big| f^\prime(z)\big|^2\) |
(D) \(8 \big| f^\prime(z)\big|^2\)
|
Select a simple analytic function and test which option is correct.
Take \(f(z)=z\) and then it is easy to check that Option (C) is the correct option.
With choice \(f(z)=z\), we get \( \big|f(z)\big|^2 = (x^2+y^2)\).
Therefore
\(\left( \frac{\partial^2 }{\partial x^2} + \frac{\partial^2 }{\partial y^2}\right) \left| f(z)\right|^2 = 4\)
Also we have \(f^\prime(z)=1\).
Hence Option (D) is the correct option.
Communicated by Sunday-Physics-Participant
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From a participant of Sunday
From a participant of Sunday Physics
Sir i got the solution as 4 ,then i have confusion why |f'(z)|^2 ?
I guess you are asking why is
I guess you are asking why is there |f'(z)|^2 in the answer?
Then the answer is as follows. You have taken f(z) =z as test case.
You computed the expression in the main stem and you get the answer 4.
Next, you must also compute the expressions in different options for the same function f(z) =z.
This give f'(z)=1. The four options now ( for this test case) become
(A) 2 (B) 3 (C) 4 (D) 8.
So option (C) is the correct option.