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Question Number 140104 by EnterUsername last updated on 04/May/21

Let z = 1 + \cos (10 π / 9) + i \sin (10 π / 9) . Then

(A) ∣ z ∣= 2 \cos (\frac{2 π}{9}) (B) \arg z = \frac{8 π}{9}

(C) ∣ z ∣= 2 \cos (\frac{4 π}{9}) (D) \arg z = \frac{5 π}{9}

Commented by EnterUsername last updated on 04/May/21

One or more answers may be correct .

Answered by MJS_new last updated on 04/May/21

\cos \frac{10 π}{9} + i \sin \frac{10 π}{9} = - \cos \frac{π}{9} - i \sin \frac{π}{9}

∣ z ∣= \sqrt{{(1 - \cos θ)}^{2} + {(- \sin θ)}^{2}} = \sqrt{2 - 2 \cos θ} =

= 2 ∣ \sin \frac{θ}{2} ∣= 2 \sin \frac{π}{18} = 2 \cos \frac{4 π}{9}

\tan (\arg z) = \frac{- \sin θ}{1 - \cos θ} = - \cot \frac{θ}{2} = - \cot \frac{π}{18} \Rightarrow

\Rightarrow \arg z = \frac{5 π}{9}

Commented by EnterUsername last updated on 04/May/21

Thank you Sir

\arg z = - \tan^{- 1} (\cot \frac{π}{18}) = \frac{π}{2} + \cot^{- 1} (\cot \frac{π}{18}) = \frac{π}{2} + \frac{π}{18} = \frac{5 π}{9}

Understood!