Math, asked by rudraku9162, 6 hours ago

Find argument of 1-√3i

Answers

Answered by ragzep0

Answer:

−2π/3

Step-by-step explanation:

Let z=−1−i

Let z=−1−i 3

Let z=−1−i 3 Then α=tan

Let z=−1−i 3 Then α=tan −1

Let z=−1−i 3 Then α=tan −1 ∣b/a∣=tan

Let z=−1−i 3 Then α=tan −1 ∣b/a∣=tan −1

Let z=−1−i 3 Then α=tan −1 ∣b/a∣=tan −1 ∣

Let z=−1−i 3 Then α=tan −1 ∣b/a∣=tan −1 ∣∣

Let z=−1−i 3 Then α=tan −1 ∣b/a∣=tan −1 ∣∣∣

Let z=−1−i 3 Then α=tan −1 ∣b/a∣=tan −1 ∣∣∣ 3

Let z=−1−i 3 Then α=tan −1 ∣b/a∣=tan −1 ∣∣∣ 3 /1

Let z=−1−i 3 Then α=tan −1 ∣b/a∣=tan −1 ∣∣∣ 3 /1 ∣

Let z=−1−i 3 Then α=tan −1 ∣b/a∣=tan −1 ∣∣∣ 3 /1 ∣∣

Let z=−1−i 3 Then α=tan −1 ∣b/a∣=tan −1 ∣∣∣ 3 /1 ∣∣∣

Let z=−1−i 3 Then α=tan −1 ∣b/a∣=tan −1 ∣∣∣ 3 /1 ∣∣∣ =π/3

Let z=−1−i 3 Then α=tan −1 ∣b/a∣=tan −1 ∣∣∣ 3 /1 ∣∣∣ =π/3Here, z is in the third quadrant.

Let z=−1−i 3 Then α=tan −1 ∣b/a∣=tan −1 ∣∣∣ 3 /1 ∣∣∣ =π/3Here, z is in the third quadrant.Therefore, argument is θ=−(π−α)=−(π−π/3)

=−2π/3

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