If $|2^z| = 1$ for a non-zero complex number $z$ then which one of the following is necessarily true.
$(A)$ $Re(z) = 0.$
$(B)$ $|z| = 1.$
$(C)$ $Re(z) = 1.$
$(D)$ No such $z$ exists.
i thinks option D) will correct because $2^z= 1$ possible only if $z=0$
On
Given that
$\vert 2^z \vert = 1, \tag 1$
we recall that
$2 = e^{\ln 2}; \tag 2$
then
$\vert (e^{\ln 2})^z \vert = 1 \Longrightarrow \vert e^{z \ln 2} \vert = 1, \tag 3$
$\vert e^{z \ln 2} \vert = 1 \Longrightarrow e^{z \ln 2} = e^{i\theta}, \theta \in \Bbb R, \tag 4$
$e^{z \ln 2} = e^{i\theta} \Longrightarrow e^{z \ln 2 - i\theta} = 1, \tag 5$
$ e^{z \ln 2 - i\theta} = 1 \Longrightarrow z \ln 2 - i\theta = 2\pi n i, \; n \in \Bbb Z; \tag 6$
thus,
$z \ln 2 = (2 \pi n + \theta)i, \; \theta \in \Bbb R, \; n \in \Bbb Z, \tag 7$
$z = \dfrac{2\pi n + \theta}{\ln 2} i; \tag 8$
so if $z \ne 0$ it is purely imaginary,
$\Re(z) = 0; \tag 9$
the correct choice is therefore (A).
We note that any real number may be expressed in the form $2 \pi n + \theta$, so in fact
$\vert 2^{ir} \vert = \vert e^{(\ln 2) ri} \vert = 1, \; r \in \Bbb R. \tag{10}$
Whenever you do complex exponentiation, always remember that $a^b=e^{b\ln a}$. Thus, we have:
$$|2^z|=|e^{z\ln 2}|=1$$
Now, the magnitude of $e^x$ is $e^{Re(x)}$, so we get:
$$e^{Re(z\ln 2)}=1\rightarrow Re(z\ln 2)=\ln 1=0\rightarrow Re(z)=\frac{0}{\ln 2}=0$$
Thus, the $Re(z)=0$ choice is correct. For example, $z=-i$, $z=i$, and $z=2i$ are all non-zero solutions to the $|2^z|=1$ equation