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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 $z=0$

Whenever you do complex exponentiation, always remember that $a^b=e^{bln a}$. Thus, we have:

$$|2^z|=|e^{zln 2}|=1$$

Now, the magnitude of $e^x$ is $e^{Re(x)}$, so we get:

$$e^{Re(zln 2)}=1rightarrow Re(zln 2)=ln 1=0rightarrow Re(z)=frac{0}{ln 2}=0$$

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

Given that

$vert 2^z vert = 1, tag 1$

we recall that

$2 = e^{ln 2}; tag 2$

then

$vert (e^{ln 2})^z vert = 1 Longrightarrow vert e^{z ln 2} vert = 1, tag 3$

$vert e^{z ln 2} vert = 1 Longrightarrow e^{z ln 2} = e^{itheta}, theta in Bbb R, tag 4$

$e^{z ln 2} = e^{itheta} Longrightarrow e^{z ln 2 - itheta} = 1, tag 5$

$ e^{z ln 2 - itheta} = 1 Longrightarrow z ln 2 - itheta = 2pi n i, ; n in Bbb Z; tag 6$

thus,

$z ln 2 = (2 pi n + theta)i, ; theta in Bbb R, ; n in Bbb Z, tag 7$

$z = dfrac{2pi n + theta}{ln 2} i; tag 8$

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

$vert 2^{ir} vert = vert e^{(ln 2) ri} vert = 1, ; r in Bbb R. tag{10}$