Question

What is the tan invers of 3i/-1-i​

Answer 1

z = 3i / (-1 - i )

z = 3i / (-1 - i ) × (-1 + i ) / (-1 + i )

z = (3i × (-1 + i )) / ((-1)² - i ²)

z = (-3i + 3i ²) / ((-1)² - i ²)

z = (-3 - 3i ) / (1 - (-1))

z = (-3 - 3i ) / 2

Note that this number lies in the third quadrant of the complex plane, where both Re(z) and Im(z) are negative. But arctan only returns angles between -π/2 and π/2. So we have

arg(z) = arctan((-3/2)/(-3/2)) - π

arg(z) = arctan(1) - π

arg(z) = π/4 - π

arg(z) = -3π/4

where I'm taking arg(z) to have a range of -π < arg(z) ≤ π.