1
$\begingroup$

" Find the points $z \in \mathbb{C}$ at which the function $g(z) = \cos(\bar{z})$ satisfies the Cauchy-Riemann equations. "

So I've written $g(z) = g(x,y) = \cos(x-iy)=\cos(x)\cosh(y)+i\sin(x)\sinh(y)$. So that $u(x,y)=\cos(x)\cosh(y)$ and $v(x,y)=\sin(x)\sinh(y)$.

From the first C-R equation I get $-\sin(x)\cosh(y) = \sin(x)\cosh(y)$ which is true when $\sin(x)\cosh(y)=0$.

Now here I thought this implied $\sin(x) = 0$ or $\cosh(y) = 0$. However the solutions give only $\sin(x) = 0$, why?

Also, when doing the second equation, you get $\cos(x)\sinh(y) = 0$, but this time the solutions split the case into $\cos(x)=0$ and $\sinh(y)=0$ as I did above.

Why this?

EDIT

Fair enough, I've found this document explaining why it is not zero, as it was said in the comments. Hyperbolic Functions

  • 3
    For real $y$ we have $\cosh(y) \geq \frac 12e^y$, so it doesn't have real roots. – 2017-01-18
  • 0
    @LeBtz More precisely, $\cosh(y)\ge 1$. – 2017-01-18
  • 0
    So $\cosh(y) =0$ only for complex arguments? Is it $z = \frac{\pi}{2}i +n\pi$ ? – 2017-01-18
  • 0
    Yes for the first part, no for the second. The roots are given by $i(\frac{\pi}{2}+n\pi), ~n\in\mathbb Z$. – 2017-01-18
  • 0
    Oh yeah it makes sense that also the period is complex, thank you – 2017-01-18

3 Answers 3

4

Now here I thought this implied $\sin(x) = 0$ or $\cosh(y) = 0$. However the solutions give only $\sin(x) = 0$, why?

$$ \cosh x = \frac{e^x + e^{-x}}2$$

So if $\cosh x = 0$, then $e^x + e^{-x} = 0$. But this can't happen because both $e^x$ and $e^{-x}$ are positive, no matter what real value $x$ has. And the sum of two positive numbers can never be zero. Therefore $\cosh x$ can never be zero for real values of $x$.

Also, when doing the second equation, you get $\cos(x)\sinh(y) = 0$, but this time the solutions split the case into $\cos(x)=0$ and $\sinh(y)=0$ as I did above.

Why this?

$$ \sinh x = \frac{e^x - e^{-x}}2$$

So if $\sinh x = 0$, then $e^x - e^{-x} = 0$. And this can happen because the difference of two positive numbers can be zero. And the solution is \begin{align*} e^x - e^{-x} &= 0\\ e^x &= \frac1{e^x}\\ e^{2x} &= 1\\ 2x &= \ln 1\\ x &= 0 \end{align*}

  • 0
    thank you, that's a perfect explanation! – 2017-01-18
1

Hint: $$\cosh^2(y)=1+\sinh^2(y)$$

What happens if $\cosh(y)=0$?

  • 0
    do we have $y = arcsinh(i)$ ? – 2017-01-18
  • 1
    @Euler Yes, but when you write complex numbers in the form $x+iy$ it is always assumed/understood that $y$ is real and hence $\sinh(y)$ is real. – 2017-01-18
1

The $\cosh$ of a real number is never zero.