Complex Analysis Solutions: Real and Imaginary Parts

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SUMMARY

The discussion focuses on finding the real and imaginary parts of complex functions, specifically for the equations w = ((e^(conjugated(z)))^2) and w = tgz. The solutions provided are: for the first equation, the real part u = (e^(x^2-y^2))cos2xy and the imaginary part v = -(e^(x^2-y^2))sin2xy; for the second equation, the real part u = (sinxcosx)/(ch^2y-sin^2x) and the imaginary part v = (shychy)/(ch^2-sin^2x). The discussion also includes hints on using LaTeX for mathematical notation.

PREREQUISITES
  • Understanding of complex numbers and their properties
  • Familiarity with exponential functions and trigonometric identities
  • Knowledge of LaTeX for mathematical formatting
  • Basic calculus concepts related to real and imaginary parts
NEXT STEPS
  • Study the properties of complex conjugates and their applications
  • Learn about the exponential form of complex numbers
  • Explore the use of LaTeX for writing mathematical expressions
  • Investigate the applications of complex functions in engineering and physics
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Mathematicians, physics students, engineers, and anyone interested in complex analysis and its applications in real-world problems.

Fabio010
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Find the real part and imaginary part of the following exercises.

1) w = ((e^(conjugated(z)))^2

2) w = tgz



Solutions:

1) u= (e^(x^2-y^2))cos2xy v= -(e^(x^2-y^2))sin2xy

2) u= (sinxcosx)/(ch^2y-sin^2x) v= (shychy)/(ch^2-sin^2x)



-------------------------------------

Attempts:
 
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Fabio010 said:
Find the real part and imaginary part of the following exercises.

1) w = ((e^(conjugated(z)))^2

2) w = tgz



Solutions:

1) u= (e^(x^2-y^2))cos2xy v= -(e^(x^2-y^2))sin2xy

2) u= (sinxcosx)/(ch^2y-sin^2x) v= (shychy)/(ch^2-sin^2x)



-------------------------------------

Attempts:


It won't take you long to learn enough LaTeX to properly write mathematics in this forum...

Hints: putting \,z:=x+iy\,\,,\,\,x,y\in\Bbb R\,:

$$(1)\,\,\left(e^{\overline z}\right)^2=\left(e^{x-iy}\right)^2=e^{2x-2iy}=e^{2x}e^{-2iy}=e^{2x}\left(\cos 2y-i\sin 2y\right)$$

$$(2)\;\;\;\tan z=\frac{\sin z}{\cos z}=\frac{e^{iz}-e^{-iz}}{2i}\cdot\frac{2}{e^{iz}+e^{-iz}}=\frac{1}{i}\frac{e^{2iz}-1}{e^{2iz}+1}=i \frac{1-e^{2iz}}{1+e^{2iz}} $$

and now you can use (1) above

DonAntonio
 

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