# For complex z showing $\cos(z+ \pi) = -\cos(z)$

As the title states, for complex $z$ I want to show $\cos(z+ \pi) = -\cos(z)$.

My first attempt was to change $\cos$ into $(e^{iz} + e^{-iz}) /2$ but then I figured using the identity $\cos(z) = \cos(x)\cosh(y)+i\sin(x)\sinh(y)$ was better since $\cos(x+ \pi)=-\cos(x)$ for real $x$. But now I'm unsure how to proceed.

• The easiest way is to use the principle of permanence, but this principle is not trivial. – lhf Sep 30 '15 at 3:02

Hint: Recall that $e^{\pi i} = e^{-\pi i} = -1$ and that $e^{u+v} = e^{u}e^{v}$ and go with your first way.

In both cases you must get the same answer. For instance, with the second one we have, for $z=x+iy$, \begin{align} \cos(z+\pi)&=\cos(x+\pi+iy)\\ &=\cos(x+\pi)\cosh y+i\sin(x+\pi)\sin y\\ &=(\cos x\cos\pi-\sin x\sin \pi)\cosh y+i(\sin x\cos \pi +\sin\pi\sin x)\sinh y\\ &=\left[(\cos x)(-1)-(\sin x)(0)\right]\cosh y+i[(\sin x)(-1)+(0)\cos x]\sinh y\\ &=-\cos x\cosh y-i\sin x\sinh y \tag 1 \end{align} Added:

Now, $$-\cos z=-(\cos x\cosh y+i\sin x\sinh y)=-\cos x\cosh y-i\sin x\sinh y\tag 2$$

By comparing ($1$) and ($2$) the identity follows.

• But isn't that -(cos$x$cosh$y$+isin$x$sinh$y$)? – Lupin Sep 30 '15 at 3:14
• Which is different from -(cos$x$cosh$y$-sin$x$sinh$y$)? – Lupin Sep 30 '15 at 3:14
• @Lupin: I don't understand what are you asking. – Ángel Mario Gallegos Sep 30 '15 at 3:15

The sleek, more complex-analytic way to do it: Let $f(z) = \cos z + \cos(z+\pi)$. Then $f$ is entire, and $f(x) = 0$ for $x \in \mathbb{R}$. Hence, by the identity theorem, $f(z) = 0$ for all $z \in \mathbb{C}$.

• That's the principle of permanence... – lhf Oct 1 '15 at 1:46

$\cos(z) = \cos(x)\cosh(y)+i\sin(x)\sinh(y)$
Then just plug in $z+\pi$ instead:
$\cos(z+\pi) = \cos(x+\pi)\cosh(y)+i\sin(x+\pi)\sinh(y)$
And observe that both $\cos(x+\pi)=-\cos(x)$ and $\sin(x+\pi)=-\sin(x)$ and you get:
$\cos(z+\pi) = -\cos(x)\cosh(y)-i\sin(x)\sinh(y) = -\cos(z)$