Can you please tell me if this question is Correct?

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In summary, the conversation discusses whether the function u(x,y) = e3x((3y-2)cos(3y)-2xsin(3y)) is holomorphic or harmonic in C and how to find its complex potential. It is concluded that the function is not holomorphic or harmonic in its current form, but can be made harmonic by replacing the 2 with a 3.
  • #1
shen07
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"Show that u(x,y) = e3x((3y-2)cos(3y)-2xsin(3y)) is holomorphic in C.

Hence Find the complex Potential f(z) such that u(x,y)=Re(f(x+iy))."

Well i think it should be harmonic because we cannot show that Re(z) is holomorphic and hence find its Complex Potential, just want you to confirm that with me.
 
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  • #2
shen07 said:
"Show that u(x,y) = e3x((3y-2)cos(3y)-2xsin(3y)) is holomorphic in C.

Hence Find the complex Potential f(z) such that u(x,y)=Re(f(x+iy))."

Well i think it should be harmonic because we cannot show that Re(z) is holomorphic and hence find its Complex Potential, just want you to confirm that with me.
I agree with you. A real-valued function cannot be holomorphic (unless it is constant). The question should be asking whether u(x,y) is harmonic, not holomorphic.
 
  • #3
Opalg said:
I agree with you. A real-valued function cannot be holomorphic (unless it is constant). The question should be asking whether u(x,y) is harmonic, not holomorphic.
Ok. Thanks..But the function also is not right then? because LaPlace's Equation does not hold from that function..
 
  • #4
shen07 said:
"Show that u(x,y) = e3x((3y-2)cos(3y)-2xsin(3y)) is holomorphic in C.

Hence Find the complex Potential f(z) such that u(x,y)=Re(f(x+iy))."

Well i think it should be harmonic because we cannot show that Re(z) is holomorphic and hence find its Complex Potential, just want you to confirm that with me.

shen07 said:
Ok. Thanks..But the function also is not right then? because LaPlace's Equation does not hold from that function..
You're right again! As it stands, that function is not harmonic. Perhaps the last 2 is a misprint for a 3. Then $u(x,y) = e^{3x}\bigl((3y-2)\cos(3y)-\mathbf{3}x\sin(3y)\bigr)$ is harmonic, being the real part of $-(3iz+2)e^{3z}.$
 
  • #5
Opalg said:
You're right again! As it stands, that function is not harmonic. Perhaps the last 2 is a misprint for a 3. Then $u(x,y) = e^{3x}\bigl((3y-2)\cos(3y)-\mathbf{3}x\sin(3y)\bigr)$ is harmonic, being the real part of $-(3iz+2)e^{3z}.$

Hey Thanks a lot for that, i would have passed hours to find that error, I am really greatful to you.
 

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