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Question

# Find all the cube root of −4√2−4√2i

A

[cos + isin ] , ,

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B

, [cos2 \pi + isin 2 \pi ] ,

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C

, ,

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D

, ,

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Solution

## The correct option is C , , Let z = −4√2−4√2i Write given complex number in polar form |z| = √(−4√2)2+(−4√2)2 = 16 tan−(α) = −4√2−4√2 = 1 α = π4 Since. complex number is in third quadrant Then, argument θ=−(π−α) = -(π−π4)= −3π4 Polar form of z = 16 [cos(−3π4)+isin(−3π4)] = 16 (cos3π4−isin3π4) Cube root of z = 1613[cos(3π4)+isin(−3π4)]13 =1613[cos(2kπ−3π4)3+isin(2kπ−3π4)3] When k = 0,1,2 When k = 0 1613 [cos(−3π4)3+isin(−3π4)3] = 1613 [cosπ4−isinπ4] When k =1 1613 [cos(2π−3π4)3+isin(2π−3π4)3] 1613 [cos5π12+isin5π12] When k = 2 1613 [cos(4π−3π4)3+isin(4π−3π4)3] 1613 [cos(13π12)+isin13π12]

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