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.

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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.

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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))#