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

Jan 19, 2017

$\left\mid {\left(3 - i\right)}^{15} \right\mid = {10}^{7} \sqrt{10}$

${\left(3 - i\right)}^{15} = 3593088 + 31417984 i$

$A r g \left({\left(3 - i\right)}^{15}\right) = {\tan}^{- 1} \left(\frac{245453}{28071}\right) \approx 1.456927$

#### Explanation:

First note that:

$\left\mid 3 - i \right\mid = \sqrt{{3}^{2} + {1}^{2}} = \sqrt{10}$

So:

$\left\mid {\left(3 - i\right)}^{15} \right\mid = {\left(\sqrt{10}\right)}^{15} = {10}^{7} \sqrt{10}$

I misread the question, so below I have calculated the value of ${\left(3 - i\right)}^{15}$ rather than the principal value of the argument, but anyway...

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

$\left(a + b i\right) = \left({a}^{2} - {b}^{2}\right) + 2 a b i$

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

${\left(3 - i\right)}^{15} = {\left(3 - i\right)}^{16} / \left(3 - i\right)$

$\textcolor{w h i t e}{{\left(3 - i\right)}^{15}} = {\left({\left({\left({\left(3 - i\right)}^{2}\right)}^{2}\right)}^{2}\right)}^{2} / \left(3 - i\right)$

$\textcolor{w h i t e}{{\left(3 - i\right)}^{15}} = {\left({\left({\left(8 - 6 i\right)}^{2}\right)}^{2}\right)}^{2} / \left(3 - i\right)$

$\textcolor{w h i t e}{{\left(3 - i\right)}^{15}} = {\left({\left(28 - 96 i\right)}^{2}\right)}^{2} / \left(3 - i\right)$

$\textcolor{w h i t e}{{\left(3 - i\right)}^{15}} = {\left(- 8432 - 5376 i\right)}^{2} / \left(3 - i\right)$

$\textcolor{w h i t e}{{\left(3 - i\right)}^{15}} = \frac{42197248 + 90660864 i}{3 - i}$

$\textcolor{w h i t e}{{\left(3 - i\right)}^{15}} = \frac{\left(42197248 + 90660864 i\right) \left(3 + i\right)}{\left(3 - i\right) \left(3 + i\right)}$

$\textcolor{w h i t e}{{\left(3 - i\right)}^{15}} = \frac{35930880 + 314179840 i}{10}$

$\textcolor{w h i t e}{{\left(3 - i\right)}^{15}} = 3593088 + 31417984 i$

Now back to the question as written...

Note that the real and imaginary parts of ${\left(3 - i\right)}^{15}$ are both positive, so it lies in Q1.

Hence the principal value of the argument is simply:

${\tan}^{- 1} \left(\frac{31417984}{3593088}\right) = {\tan}^{- 1} \left(\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{128}}} \cdot 245453}{\textcolor{red}{\cancel{\textcolor{b l a c k}{128}}} \cdot 28071}\right) = {\tan}^{- 1} \left(\frac{245453}{28071}\right) \approx 1.456927$

Jan 19, 2017

$\left\mid {\left(3 - i\right)}^{15} \right\mid = {10}^{7} \sqrt{7}$

$A r g \left({\left(3 - i\right)}^{15}\right) = 15 {\tan}^{- 1} \left(- \frac{1}{3}\right) + 2 \pi \approx 1.45693$

#### Explanation:

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

$\left\mid 3 - i \right\mid = \sqrt{{3}^{2} + {1}^{2}} = \sqrt{10}$

Hence:

$3 - i = \sqrt{10} \left(\cos \alpha + i \sin \alpha\right)$

where $\alpha = {\tan}^{- 1} \left(- \frac{1}{3}\right)$

Then:

${\left(3 - i\right)}^{15} = {\left(\sqrt{10} \left(\cos \alpha + i \sin \alpha\right)\right)}^{15}$

$\textcolor{w h i t e}{{\left(3 - i\right)}^{15}} = {\left(\sqrt{10}\right)}^{15} \left(\cos 15 \alpha + i \sin 15 \alpha\right)$

$\textcolor{w h i t e}{{\left(3 - i\right)}^{15}} = {10}^{7} \sqrt{10} \left(\cos 15 \alpha + i \sin 15 \alpha\right)$

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

$15 {\tan}^{- 1} \left(- \frac{1}{3}\right) \approx 15 \cdot \left(- 0.32175\right) = - 4.82625$

Add $2 \pi$ to this to bring it into the range $\left[- \pi , \pi\right]$:

$- 4.82625 + 2 \pi \approx - 4.82625 + 2 \cdot 3.14159 = 1.45693$