Cauchy-Riemann Equations

[djrh]

Hello, I am struggling of how to start a Cauchy-Riemann equation to show that they hold.
The equation in question is:
f(z)=cos(2zi)
I haven't came across them with trig and imaginary numbers involved before I have usually dealt with simpler equations such as f(z)=f^3 ect.
Any help would be great.

 

[Deleted member 4993]

Hello, I am struggling of how to start a Cauchy-Riemann equation to show that they hold.
The equation in question is:
f(z)=cos(2zi)
I haven't came across them with trig and imaginary numbers involved before I have usually dealt with simpler equations such as f(z)=f^3 ect.
Any help would be great.

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..... start a Cauchy-Riemann equation to show that they hold​

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[djrh]

For example for f(z)=z^3 the first step would be to substitute the z for z=x+yi giving (x+yi)^3
So for f(z)=cos(2zi) would you do the same which would leave you with:
Cos(2(x+yi)+i) (not simplified)