If |z1| = |z2| = ….. = |zn| = 1,

According to the question,

We have,

|z1| = |z2| = … = |zn| = 1

⇒ |z1|2 = |z2|2 = … = |zn|2 = 1

⇒ z1 z̅ 1= z2 z̅ 2= z3 z̅ 3= … = zn z̅ n= 1

$\Rightarrow \mathrm{z}_{1}=\frac{1}{\overline{\mathrm{z}_{1}}}, \mathrm{z}_{2}=\frac{1}{\overline{\mathrm{z}_{2}}}, \ldots, \mathrm{z}_{\mathrm{n}}=\frac{1}{\overline{\mathrm{z}_{\mathrm{n}}}}$

Now,

$\Rightarrow\left|\mathrm{z}_{1}+\mathrm{z}_{2}+\mathrm{z}_{3}+\mathrm{z}_{4}+\cdots+\mathrm{z}_{\mathrm{n}}\right|=\left|\frac{\mathrm{z}_{1} \overline{\mathrm{z}_{1}}}{\overline{\mathrm{z}_{1}}}+\frac{\mathrm{z}_{2} \overline{\mathrm{z}_{2}}}{\overline{\mathrm{z}_{2}}}+\frac{\mathrm{z}_{3} \overline{\mathrm{z}_{3}}}{\overline{\mathrm{z}_{3}}}+\cdots+\frac{\mathrm{z}_{\mathrm{n}} \overline{\mathrm{z}_{\mathrm{n}}}}{\overline{\mathrm{z}_{\mathrm{n}}}}\right|$

$=\left|\frac{1}{\overline{\mathrm{z}_{1}}}+\frac{1}{\overline{\mathrm{z}_{2}}}+\frac{1}{\overline{\mathrm{z}_{3}}}+\cdots+\frac{1}{\overline{\mathrm{z}_{n}}}\right|$

$=\left|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}+\cdots+\frac{1}{z_{n}}\right|$

$=\left|\frac{1}{\mathrm{z}_{1}}+\frac{1}{\mathrm{z}_{2}}+\cdots+\frac{1}{\mathrm{z}_{\mathrm{n}}}\right|$

Hence proved.