This one is in the second quadrant, so the principal value of the inverse tangent misses it.
Trigonometric form of a + bi is
r\ "cis"\ theta = r ( cos theta + i sin theta )
where
a = r cos theta and b = r sin theta
Squaring and adding we see
a^2 + b^2 = r^2 cos ^2 theta + r^2 sin ^2 theta = r^2
We can divide and write theta = arctan (b/a) but that's inadequate as it conflates pairs of quadrants. Let's write
theta = "arc""tan2"(b //,a )
That's deliberately the two parameter, four quadrant inverse tangent. The funky "/," is deliberate, reminding us this is two parameters and which is which. It works for a=0. If r^2=a^2+b^2 the four quadrant inverse tangent assures
a = r cos theta and b = r sin theta
In the second quadrant we have
"arc""tan2"(15 //, -3 ) = "Arc""tan"(-5) + 180^circ approx 101.3 ^circ
r = \sqrt{(-3)^2+(15)^2} = 3 sqrt{26}
-3 + 15 i = 3 sqrt{26} \ "cis" \ ( "Arc""tan"(-5) + 180^circ )
-3 + 15 i approx 3 sqrt{26} \ "cis" \ ( 101.3^circ )
