If |z_1| = |z_2| = dots = |z_n| = 1,then the | vedclass

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(C) Given $|z_k| = 1$ for $k = 1, 2, \dots, n$.Since $|z_k|^2 = z_k \overline{z_k} = 1$,we have $\overline{z_k} = \frac{1}{z_k}$.We know that $|z| = |

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$\left| \frac{1}{z_1} + \frac{1}{z_2} + \dots + \frac{1}{z_n} \right|$

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(C) Given $|z_k| = 1$ for $k = 1, 2, \dots, n$.Since $|z_k|^2 = z_k \overline{z_k} = 1$,we have $\overline{z_k} = \frac{1}{z_k}$.We know that $|z| = |\overline{z}|$.Therefore,$|z_1 + z_2 + \dots + z_n| = |\overline{z_1 + z_2 + \dots + z_n}|$.Using the property of conjugates,$|\overline{z_1} + \overline{z_2} + \dots + \overline{z_n}| = \left| \frac{1}{z_1} + \frac{1}{z_2} + \dots + \frac{1}{z_n} \right|$.Thus,$|z_1 + z_2 + \dots + z_n| = \left| \frac{1}{z_1} + \frac{1}{z_2} + \dots + \frac{1}{z_n} \right|$.

If $|z_1| = |z_2| = \dots = |z_n| = 1$,then the value of $|z_1 + z_2 + z_3 + \dots + z_n|$ is equal to:

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