JEE MAIN - Mathematics (2019 - 8th April Evening Slot - No. 3)
If $$z = {{\sqrt 3 } \over 2} + {i \over 2}\left( {i = \sqrt { - 1} } \right)$$,
then (1 + iz + z5 + iz8)9 is equal to :
then (1 + iz + z5 + iz8)9 is equal to :
1
–1
0
(-1 + 2i)9
Explanation
$$z = {{\sqrt 3 } \over 2} + {i \over 2}$$
$$ \Rightarrow $$ z = $$\cos {\pi \over 6}$$ + i $$\sin {\pi \over 6}$$
$$ \Rightarrow $$ z = $${e^{i{\pi \over 6}}}$$_8th_April_Evening_Slot_en_3_2.png)
$$ \Rightarrow $$ z = $$\cos {\pi \over 6}$$ + i $$\sin {\pi \over 6}$$
$$ \Rightarrow $$ z = $${e^{i{\pi \over 6}}}$$
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