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 ) #
