How do you calculate #abs(-3+2i)#?

In general, for a complex number #z=a+b"i"#, we define the modulus of #z# in the following way,

#abs(z) = sqrt{a^2+b^2}#.

Applying this to your specific case gives

#abs(-3+2"i") = sqrt{(-3^2)+(2^2)}#.

You've put this question under the trigonometric form of a complex number, so I'll offer some explanation for how the modulus of a complex number relates to its trigonometric form.

Any number on the unit circle in the complex plane can be represented on can be represented as the complex number #w# where #w = cos(theta)+"i"sin(theta)#.

In order to represent points in the complex plane not just lying on the unit circle, the number needs to be scaled by the distance from the origin the point is.

As can be seen from the image above (and the Pythagorean theorem) the modulus of #z# is simply the distance from the point in the plane #z# represents to the origin. This is the scaling factor required.

So, for any complex number,

#z=a+b"i"#,

we can write #z# in the form,

#z=abs(z)(cos(theta)+"i"sin(theta))#,

where #abs(z) = sqrt{a^2+b^2}#, and #theta# (#-pi< theta \leq pi#) is the argument of z calculated from the angle the point #(a,b)# makes with the positive real axis.

So to represent #-3+2i# in polar form, you would calculate its argument #theta# from the unit circle and then write,

#z=-3+2i=sqrt{13}(cos(theta)+"i"sin(theta))#