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Trigonometric Equations

Find all the ...

Question

Find all the cube root of $- 4 \sqrt{2} - 4 \sqrt{2} i$

A

[cos + isin ] , ,

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B

, ,

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C

, ,

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D

, ,

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Solution

The correct option is C
, ,

Let z = $- 4 \sqrt{2} - 4 \sqrt{2} i$

Write given complex number in polar form

$| z |$ = $\sqrt{(- 4 \sqrt{2})^{2} + (- 4 \sqrt{2})^{2}}$ = 16

$t a n^{-} (α)$ = $\frac{- 4 \sqrt{2}}{- 4 \sqrt{2}}$ = 1

α = $\frac{π}{4}$

Since. complex number is in third quadrant

Then, argument $θ = - (π - α)$ = - $(π - \frac{π}{4})$ = $\frac{- 3 π}{4}$

Polar form of z = 16 $[c o s (\frac{- 3 π}{4}) + i s i n (\frac{- 3 π}{4})]$

= 16 $(c o s \frac{3 π}{4} - i s i n \frac{3 π}{4})$

Cube root of z = $16^{\frac{1}{3}} {[c o s (\frac{3 π}{4}) + i s i n (\frac{- 3 π}{4})]}^{\frac{1}{3}}$

$= 16^{\frac{1}{3}} [c o s \frac{(2 k π - \frac{3 π}{4})}{3} + i s i n \frac{(2 k π - \frac{3 π}{4})}{3}]$

When k = 0,1,2

When k = 0

$16^{\frac{1}{3}}$ $[c o s \frac{(\frac{- 3 π}{4})}{3} + i s i n \frac{(\frac{- 3 π}{4})}{3}]$

= $16^{\frac{1}{3}}$ $[c o s \frac{π}{4} - i s i n \frac{π}{4}]$

When k =1

$16^{\frac{1}{3}}$ $[c o s \frac{(2 π - \frac{3 π}{4})}{3} + i s i n \frac{(2 π - \frac{3 π}{4})}{3}]$

$16^{\frac{1}{3}}$ $[c o s \frac{5 π}{12} + i s i n \frac{5 π}{12}]$

When k = 2

$16^{\frac{1}{3}}$ $[c o s \frac{(4 π - \frac{3 π}{4})}{3} + i s i n \frac{(4 π - \frac{3 π}{4})}{3}]$

$16^{\frac{1}{3}}$ $[c o s (\frac{13 π}{12}) + i s i n \frac{13 π}{12}]$

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Find all the cube root of $- 4 \sqrt{2} - 4 \sqrt{2} i$

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