Solve $3= 3^{z}.$

Question

I'm stuck in solving this strange and beautiful formula : $3= 3^{z}$ since it says 'Solve' and not 'Prove'

Also i really don't understand what does it means by saying $3^{z}$? Will $3^z$ form a set ?

I will be very grateful if you can help me or even give me a little hint to make a headway with this problem ...

To be clear, find all $z \in \left\{ 3 = 3^z \right\}.$

Answer 1

Since $z$ is a complex number then $z=x+iy$ where $x,y \in \mathbb{R}$

Also Since $0,1 \in \mathbb{R}$ then $z=x+iy=(1)+i(0)=1$

Thus $1 \in \mathbb{Z}$ And $3^{1}=3 \in 3^{z}$

Answer 2

Well, if the question is asking to find all $z$ such that

$$3^z = 3$$

Then

$$\begin{aligned} 3^z &= 3e^{2 k \pi i } \\ z \log 3 &=\log 3 + 2 k \pi i \\ \end{aligned}$$ $$\bbox[5px, border: 1pt solid blue]{z = 1 + \frac{2 k \pi i}{\log 3}, \quad k \in \mathbb{Z}.} $$

There are infinitely many solutions, and they all lie on the line $\text{Re }z=1.$