What is the value of $\cosh(\sqrt{i})$?

Question

I am puzzling about the value of

$$\cosh(\sqrt{i})$$

I know that

$$\sqrt{i} = \sqrt{\frac{1}{2}}+i\sqrt{\frac{1}{2}}$$

But how to go on with that? Are there also multiple values?

Thank you all in advance!

Answer 1

You can use the fact that: $$\cosh(x)=\frac{e^x+e^{-x}}{2}$$ And $$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$ Also make use of compound angle formula

Answer 2

Use the definition $\cosh(x) = \frac{1}{2}(e^{x}+e^{-x})$ and the identity $\cosh(x+y) = \cosh(x)\cosh(y)+\sinh(x)\cosh(y)$ so

$$\cosh(\sqrt{i}) = \cosh(\sqrt{1/2})\cosh(i\sqrt{1/2}) + \sinh(\sqrt{1/2})\sinh(i\sqrt{1/2})$$ $$= \cosh(\sqrt{1/2})\cos(\sqrt{1/2}) + i\sinh(\sqrt{1/2})\sin(\sqrt{1/2})$$

Answer 3

Note that there are two possible values for $\sqrt{i}$, namely, $\pm\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)$. But since $\cosh$ is an even function, it doesn't matter which we choose. So we will work with finding $\cosh\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)$.

One easily verifies that $\cosh(a+bi)=\cosh a \cos b + i\sinh a\sin b$. Thus

$$\cosh\left(\tfrac{1}{\sqrt{2}}+\tfrac{i}{\sqrt{2}}\right) = \boxed{\cosh \tfrac{1}{\sqrt{2}} \cos \tfrac{1}{\sqrt{2}} + i\sinh \tfrac{1}{\sqrt{2}}\sin \tfrac{1}{\sqrt{2}}} \approx 0.9584 + 0.4986i.$$

---

Addendum: The rules for handling $f(-t)$ for trig functions $f$ have analogs for handling expressions like $f(it)$; moreover, the same rules apply to hyperbolic and trig functions alike. Let's explore them briefly.

We will distinguish between "c" and "s" functions. The familiar rules are:

• A factor of $-1$ can be pulled out of c-functions and disappears
• A factor of $-1$ can be pulled out of s-functions and remains

So: $\cos(-t)=\cos t$, and $\cosh(-t)=\cosh t$; $\sin(-t)=-\sin t$, and $\sinh(-t)=-\sinh t$.

The new rules are only slightly different:

• A factor of $i$ can be pulled out of c-functions and disappears, transforming the function
• A factor of $i$ can be pulled out of s-functions and remains, transforming the function

(here, "transforming" means changing trig to hyperbolic and vice versa).

So: $\cos(it)=\cosh t$, and $\cosh(it)=\cos t$; $\sin(it)=i\sinh t$, and $\sinh(it)=i\sin t$.

Now it is easy to see what we said above:

$$\cosh(a+bi)=\cosh a\cosh(bi) + \sinh a\sinh(bi)$$ $$= (\cosh a)( \cos b) + (\sinh a) (i\sin b)$$ $$= \cosh a \cos b + i\sinh a\sin b$$ as desired.