How do I show $\arccos(4)$ in form $a+bi$?

Question

How do I show $\arccos(4)$ is ${2\pi k}$ $\pm$ $i\operatorname{arcosh}(4)$?

I am getting $\pm$ $iln(4+\sqrt{15}$)

When I just use $\cos(z)$ =$\cosh(iz)$, I get: $-i\operatorname{arcosh}(4)$.

I’m lost! If someone could type out the full proof, it'd be greatly appreciated. I'm not seeing it from the answers below.

Thanks.

Edit: I follow the answers on the whole, but cannot seem to derive why arcosh has a prefix of $\pm$. If someone could show a derivation for this I'd be grateful. From function theory I'm expecting only the one positive root, not the -ve one. That said, I get the $\pm$ for the $\ln(x)$ version so I believe it should be there, but need a proof! :) Thanks.

Answer 1

$\arccos 4 = a+bi$

$4 = \cos(a+bi) = \cos a \cos bi - \sin a \sin bi = \cos a \cosh b - i\sin a \sinh b$

By comparing real and imaginary parts,

$\cos a \cosh b = 4$ and $\sin a \sinh b = 0$

Where $a, b$ are both real.

One possible solution to the second equation is $\sinh b =0 \implies b=0$, but that would make the first equation $\cos a = 4$, which has no real solution. So disregard this, i.e. $b \neq 0$.

Which leaves $\sin a =0 \implies a = n\pi, n \in \mathbb{Z}$.

In the first equation that makes $\pm \cosh b = 4$. Since the hyperbolic cosine for real values is non-negative only accept the positive case.

So $a = 2k\pi, k \in \mathbb{Z}$ (only even multiples of $\pi$ give a positive cosine) and $\cosh b = 4 \implies b = \pm \cosh^{-1} 4$ (since the hyperbolic cosine is an even function).

Thus we get the solution $\arccos 4 = 2k\pi \pm i\cosh^{-1} 4, k \in \mathbb{Z}$.

Note that you can also express the $\cosh^{-1} 4$ part in terms of logarithms if you wish. I believe this is what you did. To be clear $\cosh^{-1} 4 = \ln (4 + \sqrt {4^2 - 1}) = \ln (4 + \sqrt{15})$ , which means you can also write $\arccos 4 = 2k\pi \pm i \ln (4 + \sqrt{15}), k \in \mathbb{Z}$. This is an equivalent answer.

Answer 2

If we use $e^{i\theta}=\cos{(\theta)}+i\sin{(\theta)}$, we can show that $\cos{(\theta)}=\frac{e^{i\theta}+e^{-i\theta}}{2}$ or $\cos{(\theta)}=\cosh{(i\theta)}$.

$$ \begin{aligned} \cosh{(\pm\cosh^{-1}{(4)})}&=4\\ \cosh{(i(2\pi n\pm i\cosh^{-1}{(4))})}&=4\\ \\ \cos{(2\pi n\pm i\cosh^{-1}{(4)})}&=4 \end{aligned} $$

however... i do not know the range of $\cos^{-1}{(x)}$ in complex plane

Answer 3

The point is that

$$\cosh(x +i 2 \pi k) = \cosh(x) = \cosh(-x) $$

for $k$ integer.

Answer 4

Let $z = \arccos(4)$ and use the identity $\cos(iw)=\cosh(w)$ to express,

$$\cos(z) = \cos(2\pi n\pm z ) =\cos[i(-i2\pi n \mp iz )]=\cosh(-i2\pi n \mp iz ) = 4$$

or,

$$-i2n\pi \mp iz = \text{arccosh}(4)$$

which leads to,

$$z = \arccos(4) = 2\pi k \pm i\>\text{arccosh}(4) $$