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[SOLVED] Trig complex z\cos z

dwsmith

Well-known member
Feb 1, 2012
1,673
$z\cos z$

Let $z = x + yi$.
Then $f(z) = (x + yi)\cos (x + yi)$.
By the addition rule for cosine and the identities $\cos yi = \cosh y$ and $-i\sin yi = \sinh y\Leftrightarrow \sin yi = i\sinh y$, we have that $\cos (x + yi) = \cos x\cosh y + i\sin x\sinh y$.
So
$$
f(z) = z\cos z = x\cos x\cosh y - y\sin x\sinh y + i(x\sin x\sinh y + y\cos x\cosh y).
$$
Then
$$
u(x,y) = x\cos x\cosh y - y\sin x\sinh y\quad\text{and}\quad
v(x,y) = y\cos x\cosh y + x\sin x\sinh y,
$$

I am trying to verify the CR equations but there is a negative sign difference. There has to be an error in my algebra but I can't find it. What is wrong with the above?
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
$z\cos z$

Let $z = x + yi$.
Then $f(z) = (x + yi)\cos (x + yi)$.
By the addition rule for cosine and the identities $\cos yi = \cosh y$ and $-i\sin yi = \sinh y\Leftrightarrow \sin yi = i\sinh y$, we have that $\cos (x + yi) = \cos x\cosh y\,{\color{red}-}\,i\sin x\sinh y$.
So
$$
f(z) = z\cos z = x\cos x\cosh y\,{\color{red}+}\, y\sin x\sinh y + i({\color{red}-}\,x\sin x\sinh y + y\cos x\cosh y).
$$
Then
$$
u(x,y) = x\cos x\cosh y\,{\color{red}+}\, y\sin x\sinh y\quad\text{and}\quad
v(x,y) = y\cos x\cosh y\,{\color{red}-}\, x\sin x\sinh y,
$$

I am trying to verify the CR equations but there is a negative sign difference. There has to be an error in my algebra but I can't find it. What is wrong with the above?
You had a sign error. See all my changes in red.
 

dwsmith

Well-known member
Feb 1, 2012
1,673

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Bad admin for missing the solved tag in the title for before you posted.:confused:
I clicked on the link from the home page, and it didn't show a solved tag. :p