What kind of number?

As mentioned in Michael Boyd’s solution, the Gelfond-Schneider theorem states that $a^b$ is transcendental if $a$ is algebraic and $b$ is algebraic and irrational. For the purposes of this theorem, nonreal numbers are considered irrational. The number $5+i$ is algebraic as it is a root of the equation $(x-5+i)(x-5-i) = x^2-10x+26 = 0$ . And the number $i$ is algebraic as it is a root of $x^2+1 =0$ and is irrational as it is not a real number. Therefore, $(5+i)^i$ is transcendental.

Before starting, I'd like to give credit to Kaan Berk Erdogmus for finding an error that I have now fixed.

The Gelfond–Schneider theorem says that for any $a^b$ , where a is an algebraic number and b is irrational algebraic, $a^b$ must be transcendental. This is, in my opinion, the simplest way to determine if the number is transcendental. We can conclude that $(5+i)$ is algebraic, since it can be the root of the polynomial $f(x)=x^2-10x+26$ . However, there's another way to confirm the "trancendence" (if that's a word) of the expression. Let's move to the next method.

For this method, we should remember that $ln(z)=ln|z|+iArg(z)$ , for all $z=a+bi$ , where $|z|>0$ .

Next, we can set $5+i=z$ , and rewrite the original expression as $z^i$ . You could follow this with raising e to the natural log of $z^i$ : $e^{ln(z^i)}$

This can be rewritten again as $e^{i*ln(z)}$ , and now we have $ln(z)$ alone. Next, recall the method for evaluating the natural log of complex numbers mentioned above: $ln(5+i) = ln|5+i| + iArg(5+i)$

This can again be rewritten as $ln(\sqrt{26})+i*tan(1/5)$ so when we plug $ln(z)$ back into $e^{i*ln(z)}$ , we have: $e^{i*(ln(\sqrt{26})+i*tan(1/5)}$ which can be simplified to be $e^{(i*ln(\sqrt{26})-tan(1/5)}$ , and because $e^{c-d}=\frac{e^c}{e^d}$ , we can reduce our expression to $\frac{e^{i*ln\sqrt{26}}}{e^{tan(1/5)}}$

By the Lindemann–Weierstrass theorem, ln(a) is transcendental if a is algebraic and not equal to 0 or 1. $\sqrt{26}$ is algebraic because it is a root of the polynomial $f(x)=x^2-26$ . Additionally, the Lindemann–Weierstrass theorem states that all of the trigonometric functions $sin(a), cos(a),$ and $tan(a)$ are transcendental if $a$ is nonzero and alegebraic. This means $tan(1/5)$ is transcendental.

Now, for the end... If we take the natural log of our expression, we're left with: $ln(e^i)+ln(e^{ln(\sqrt{26})})-ln(e^{tan(1/5)})$ which can be simplified to:

$i+ln(\sqrt{26})-tan(\frac{1}{5})$ , and since $ln(\sqrt{26})$ and $tan(\frac{1}{5})$ are both transcendental and $ln(\sqrt{26})-tan(\frac{1}{5}) \neq 0$ , the expression must be transcendental. We have now confirmed that $(5+i)^i$ is transcendental in two ways.