Let $ z=\frac{(2 \sqrt{3}+2 i)^{8}}{(1-i)^{6}}+\frac{(1+i)^{6}}{(2 \sqrt{3}-2 i)^{8}} . $ Let $ \theta $ be the argument of $ z $ such that $ \theta \in(-\pi, \pi] $ then $ 4 \sin \theta $ is equal to

Question

Let $ z=\frac{(2 \sqrt{3}+2 i)^{8}}{(1-i)^{6}}+\frac{(1+i)^{6}}{(2 \sqrt{3}-2 i)^{8}} . $ Let $ \theta $ be the argument of $ z $ such that $ \theta \in(-\pi, \pi] $ then $ 4 \sin \theta $ is equal to,

MARKS App · Accepted Answer

z= z 1 8 z ̄ 2 6 + z 2 6 z ̄ 1 8 = | z 1 | 16 + | z 2 | 12 | z ̄ 2 | 6 ( z ̄ 1 ) 8 arg(z)=– 8 arg ( z ̄ 1 ) – 6arg ( z ̄ 2 ) +2kπ, k∈I = 8 arg (z 1 )+6 arg (z 2 )+2kπ = 8. π 6 +6. π 4 +2kπ= 4 π 3 + 3 π 2 –2π= 5 π 6 ∴ 4 sinθ=4 sin 5 π 6 = 2

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