How is Phase Shift of an Op-Amp calculated using Polar Coordinates

I have a textbook (Op-Amp: Characteristics & Applications by Robert G. Irvine, 1981) and its Frequency response chapter started by discussing the Internal LP filters between the Input Differential Amplifier stage, the Voltage Level Shifting Stage, and the Push-Pull Output Stage.

It explains that the Capacitor will create a -45 degree phase shift at it's corner frequency (where attenuation is -3dB). and corner frequency is 1/(2*pi()RC).

It says that if the Frequency on the input of the Op-Amp is much less than the pole Frequency then the phase shift is negligible but uses the equation:

RC=159,000-ohms*0.1uF=15.9-milliSeconds,

pole F: 1/(2*pi()*.0159)-1/0.1=10-Hertz

1. When F=0.1-Hz Vout/Vin=10/(10+j(0.1))= 1 @ -0.57-degrees

2. When F=10-Hz Vout/Vin=10/(10+j(10))= 10/(14.141 @ +45 Degrees) = 0.707 @ -45-degrees

3. When F=1,000-Hz Vout/Vin=10/(10+j(1000))= 10/(1000.05 @ +89.43 Degrees) = 0.01 @ -89.43-degrees

4. When F=10,000-Hz Vout/Vin=10/(10+j(10,000))= 0.001 @ -89.94-degrees

I can use Euler's formula to transpose Polar coordinates to rectangular, but the textbook doesn't give any explanation as to how, say, "10+j10" becomes 14.141 @ -45-degrees.

or how does "10+j1000" become 1000.01 @ +89.43 degrees?

Basically how do you calculate phase shift due to a a capacitor as a function of it's frequency in polar and rectangular coordinates?