Question

How do you find modulus and principal value for the argument of `(3 - i)^15`?

Answer 1

`abs((3-i)^15) = 10^7sqrt(10)`

`(3-i)^15 = 3593088 + 31417984i`

`Arg((3-i)^15) = tan^(-1) (245453/28071) ~~ 1.456927`

Explanation:

First note that:

`abs(3-i) = sqrt(3^2+1^2) = sqrt(10)`

So:

`abs((3-i)^15) = (sqrt(10))^15 = 10^7 sqrt(10)`

I misread the question, so below I have calculated the value of `(3-i)^15` rather than the principal value of the argument, but anyway...

Note also that we can square a complex number like this:

`(a+bi) = (a^2-b^2) + 2abi`

To calculate the value, I think it may be easiest to square `(3-i)` four times then divide by `(3-i)` as follows:

`(3-i)^15 = (3-i)^16/(3-i)`

`color(white)((3-i)^15) = ((((3-i)^2)^2)^2)^2/(3-i)`

`color(white)((3-i)^15) = (((8-6i)^2)^2)^2/(3-i)`

`color(white)((3-i)^15) = ((28-96i)^2)^2/(3-i)`

`color(white)((3-i)^15) = (-8432-5376i)^2/(3-i)`

`color(white)((3-i)^15) = (42197248+90660864i)/(3-i)`

`color(white)((3-i)^15) = ((42197248+90660864i)(3+i))/((3-i)(3+i))`

`color(white)((3-i)^15) = (35930880 + 314179840i)/10`

`color(white)((3-i)^15) = 3593088 + 31417984i`

Now back to the question as written...

Note that the real and imaginary parts of `(3-i)^15` are both positive, so it lies in Q1.

Hence the principal value of the argument is simply:

`tan^(-1) (31417984/3593088) = tan^(-1) ((color(red)(cancel(color(black)(128)))*245453)/(color(red)(cancel(color(black)(128)))*28071))= tan^(-1) (245453/28071) ~~ 1.456927`

Answer 2

`abs((3-i)^15) = 10^7sqrt(7)`

`Arg((3-i)^15) = 15 tan^(-1) (-1/3) + 2pi ~~ 1.45693`

Explanation:

OK - Let us de Moivre's theorem directly...

`abs(3-i) = sqrt(3^2+1^2) = sqrt(10)`

Hence:

`3-i = sqrt(10)(cos alpha + i sin alpha)`

where `alpha = tan^(-1) (-1/3)`

Then:

`(3-i)^15 = (sqrt(10)(cos alpha + i sin alpha))^15`

`color(white)((3-i)^15) = (sqrt(10))^15 (cos 15alpha + i sin 15 alpha)`

`color(white)((3-i)^15) = 10^7sqrt(10) (cos 15alpha + i sin 15 alpha)`

So we want to find the canonical representation of the angle:

`15 tan^(-1)(-1/3) ~~ 15 * (-0.32175) = -4.82625`

Add `2pi` to this to bring it into the range `[-pi, pi]`:

`-4.82625+2pi ~~ -4.82625 + 2*3.14159 = 1.45693`