If | $$e^{e^z}$$ | = 1 for a complex number z = x + iy, x, y ∈ R,then.

which of thefollowing is true ?

Correct option is B

​$$\text{The given condition is } \left| e^{e^z} \right| = 1, \text{ where } z = x + iy. \\[10pt]\textbf{Step 1: Expand } e^{e^z} \\[10pt]\text{We can write:} \\[10pt]e^{e^z} = e^{e^{x+iy}} = e^{e^x e^{iy}}. \\[10pt]=e^{e^x \cos y + e^x i \sin y} = e^{e^x \cos y} \cdot e^{e^x i \sin y} \\[10pt]\text{The modulus of } e^{e^z} \text{ is:} \\[10pt]\left| e^{e^z} \right| = \left| e^{e^x \cos y} \cdot e^{e^x i \sin y} \right| = \\[10pt]\left| e^{e^x \cos y} \right| \cdot \left| e^{e^x i \sin y} \right| \\[10pt]\text{To satisfy the given condition:} \\[10pt]\left| e^{e^z} \right| = \left| e^{e^x \cos y} \cdot e^{e^x i \sin y} \right| = 1 \\[10pt]\text{As, } | e^{i\theta} | = 1, \\[10pt]\left| e^{e^x \cos y} \right| = 1 \ \textit{is required.} \\[10pt]\implies \cos y = 0 \implies y = \frac{(2n+1)\pi}{2} \\[10pt]\text{Hence, } \textbf{Option B} \text{ is correct.}$$​​