Question 1170517
If  you draw it out on a coordinate grid, using the x-axis for reals and the y-axis for imaginary parts you will put a point at  (5,-12).

{{{ drawing(400,400,-2,16,-16,2,grid(0), line(0,0,5,-12), locate(4,-12.5,"z(5,-12)"), locate(1,-1,theta), green(line(5,0,5,-12))

) }}}


Now:
{{{ theta = tan^-1(-12/5) = -67.38^o }}} 

And the modulus (magnitude or radius for polar form) is equal to the hypotenuse of the drawn triangle:
{{{ sqrt(5^2 + (-12)^2) = 13 }}}

Putting these together:
{{{ highlight( z = 13 (cos(-67.38^o) + i * sin(-67.38^o)) ) }}}<br>

Noting that -67.38 degrees (CW rotation of 67.38 degrees) is the same as a CCW rotation of 292.62 degrees, you can write:
{{{ highlight( z = 13 (cos(292.62^o) + i * sin(292.62^o)) ) }}}

While the negative angle might be a little more common, both answers should be acceptable.