If -1+ √ − 3 =r e i θ , then θ is equal to Options: A . &nbsp π 3 B . &nbsp - π 3 C . &nbsp 2 π 3 D . &nbsp - 2 π 3

Answer: Option C : C Here -1+ √ − 3 =r e i θ ⇒ -1+ i √ 3 =r e i θ =rcos θ =-1 And rsin θ = √ 3 Hence tan θ =- √ 3 ⇒ tan θ =tan 2 π 3 .Hence θ = 2 π 3

If -1+√−3=reiθ, then θ is equal to

Question -

If -1+ $\sqrt{- 3}$ =r $e^{i θ}$ , then $θ$ is equal to

Options:

A .

\frac{π}{3}

B . -

\frac{π}{3}

C .

\frac{2 π}{3}

D . -

\frac{2 π}{3}

Answer: Option C
:
C

Here -1+ $\sqrt{- 3}$ =r $e^{i θ}$ $\Rightarrow$ -1+ $i \sqrt{3}$ =r $e^{i θ}$

=rcos $θ$ =-1 And rsin $θ$ = $\sqrt{3}$

Hence tan $θ$ =- $\sqrt{3}$ $\Rightarrow$ tan $θ$ =tan $\frac{2 π}{3}$ .Hence $θ$ = $\frac{2 π}{3}$

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