Question

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

Answer 1

`abs(-3+2"i") = sqrt{13}`

Explanation:

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))`