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Q.

If $z = e^{i 2 π / 5}$ then $1 + z + z^{2} + z^{3} + 5 z^{4} + 4 z^{5} + 4 z^{6} + 4 z^{7} + 4 z^{8} + 5 z^{9} =$ ________

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a

$- 4 z^{2}$

b

${4z}^{3}$

c

$5 z^{4}$

d

0

answer is C.

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Detailed Solution

$\begin{array}{l} z = e^{i 2 π / 5} \Rightarrow z^{5} = e^{i 2 π} = 1 + i (0) = 1 \\ (1 + z + z^{2} + z^{3} + z^{4}) + 4 z^{4} + 4 z^{5} + 4 z^{6} + 4 z^{7} + 4 z^{6} + 5 z^{9} \\ = \frac{1 - z^{5}}{1 - z} + 4 z^{4} (1 + z + z^{2} + z^{3} + z^{4}) + 5 z^{5} \cdot z^{4} \\ = 0 + 4 z^{4} (\frac{1 - z^{5}}{1 - z}) + 5 z^{4} = 5 z^{4} \end{array}$

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