# Convert all complex numbers to trigonometric form and then simplify the expression? Write the answer in standard form.

## ((2 + 2i)^5(-3+i)^3)/((sqrt3 +i)^10

May 14, 2018

$\frac{{\left(2 + 2 i\right)}^{5} {\left(- \sqrt{3} + i\right)}^{3}}{\sqrt{3} + i} ^ 10$

$= \frac{\sqrt{3} - 1}{2} + \frac{\sqrt{3} + 1}{2} \setminus i$

#### Explanation:

As anyone who reads my answers may have noticed, my pet peeve is every trig problem involves a 30/60/90 or 45/45/90 triangle. This one has both, but $- 3 + i$ is neither.

I'm going to go out on a limb and guess the question in the book actually read:

Use trigonometric form to simplify

$\frac{{\left(2 + 2 i\right)}^{5} {\left(- \sqrt{3} + i\right)}^{3}}{\sqrt{3} + i} ^ 10$

because this way would only involve the Two Tired Triangles of Trig.

Let's convert to trigonometric form, which is just polar form written

$r \setminus \textrm{c i s} \setminus \theta = r \left(\setminus \cos \theta + i \setminus \sin \theta\right)$

Then by De Moivre's Thorem

${\left(r \setminus \textrm{c i s} \setminus \theta\right)}^{n} = {r}^{n} \setminus \textrm{c i s} \left(n \theta\right)$

Let's convert each factor.

$| 2 + 2 i | = \sqrt{{2}^{2} + {2}^{2}} = 2 \sqrt{2}$

$\angle 2 + 2 i = {45}^{\circ}$

$2 + 2 i = 2 \setminus \sqrt{2} \setminus \textrm{c i s} \setminus {45}^{\circ}$

$| - \sqrt{3} + i | = \sqrt{{\left(- \sqrt{3}\right)}^{2} + {1}^{2}} = 2$

$\angle \left(- \sqrt{3} + i\right) = {150}^{\circ}$

That's the 30/60/90 triangle in the second quadrant whose cosine is bigger (in magnitude) than its sine.

$- \sqrt{3} + i = 2 \setminus \textrm{c i s} \setminus {150}^{\circ}$

Similarly,

$\setminus \sqrt{3} + i = 2 \setminus \textrm{c i s} \setminus {30}^{\circ}$

Now,

$\frac{{\left(2 + 2 i\right)}^{5} {\left(- \sqrt{3} + i\right)}^{3}}{\sqrt{3} + i} ^ 10$

$= \frac{{\left(2 \setminus \sqrt{2} \setminus \textrm{c i s} \setminus {45}^{\circ}\right)}^{5} {\left(2 \setminus \textrm{c i s} \setminus {150}^{\circ}\right)}^{3}}{{\left(2 \setminus \textrm{c i s} \setminus {30}^{\circ}\right)}^{10}}$

$= \frac{{\left(2 \setminus \sqrt{2}\right)}^{5} \setminus \textrm{c i s} \setminus \left(5 \cdot {45}^{\circ}\right) \left({2}^{3} \setminus \textrm{c i s} \setminus \left(3 \cdot {150}^{\circ}\right)\right)}{{2}^{10} \setminus \textrm{c i s} \setminus \left(10 \cdot {30}^{\circ}\right)}$

= {2^5 \sqrt{2^5} \ 2^3 }/ 2^{10} {\ text{cis}\ 225^circ \ text{cis}\ 450^circ }/{\ text{cis}\ 300^\circ }

$= \sqrt{2} \setminus \textrm{c i s} \setminus \left({225}^{\circ} + {450}^{\circ} - {300}^{\circ}\right)$

$= \setminus \sqrt{2} \setminus \textrm{c i s} \setminus {375}^{\circ}$

$= \setminus \sqrt{2} \setminus \textrm{c i s} \setminus {15}^{\circ}$

I don't want to deal with the trig functions of ${15}^{\circ} ,$ which are best gotten from the difference angle formulas with ${15}^{\circ} = {45}^{\circ} - {30}^{\circ} ,$ our tired triangles. We can avoid it by redoing the problem, cleverly leaving one factor of $2 + 2 i$ till the end:

$\frac{{\left(2 + 2 i\right)}^{5} {\left(- \sqrt{3} + i\right)}^{3}}{\sqrt{3} + i} ^ 10$

$= \left(2 + 2 i\right) \frac{{\left(2 \setminus \sqrt{2}\right)}^{4} \setminus \textrm{c i s} \setminus \left(4 \cdot {45}^{\circ}\right) \left({2}^{3} \setminus \textrm{c i s} \setminus \left(3 \cdot {150}^{\circ}\right)\right)}{{2}^{10} \setminus \textrm{c i s} \setminus \left(10 \cdot {30}^{\circ}\right)}$

$= \left(2 + 2 i\right) \frac{{2}^{4} \setminus \sqrt{{2}^{4}} \setminus {2}^{3}}{2} ^ \left\{10\right\} \frac{\setminus \textrm{c i s} \setminus {180}^{\circ} \setminus \textrm{c i s} \setminus {450}^{\circ}}{\setminus \textrm{c i s} \setminus {300}^{\setminus} \circ}$

= (2+2i)/2 \ text{cis}\ 30^circ

$= \left(1 + i\right) \cdot \frac{\setminus \sqrt{3} + i}{2}$

$= \frac{\sqrt{3} - 1}{2} + \frac{\sqrt{3} + 1}{2} \setminus i$

May 14, 2018

$\frac{11 + 2 \sqrt{3}}{4} + \frac{11 \sqrt{3} - 2}{4} i$

#### Explanation:

In another answer to this question I guessed there was a typo in this question and that $- 3$ was supposed to be $- \sqrt{3}$. I've been assured in a comment that that's not the case, that the question is correct as written.

I won't repeat how we determined

$2 + 2 i = 2 \sqrt{2} \setminus \setminus \textrm{c i s} \setminus {45}^{\circ}$

$\sqrt{3} + i = 2 \setminus \setminus \textrm{c i s} \setminus {30}^{\circ}$

But now we have to convert $- 3 + i$ to trigonometric form. We can do it, but since it's not one of Trig's preferred triangles, it's a bit more awkward.

$| - 3 + i | = \sqrt{{3}^{2} + {1}^{2}} = \sqrt{10}$

We're in the second quadrant and the principal value of the inverse tangent is the fourth quadrant.

$\setminus \angle \left(- 3 + i\right) = \setminus \textrm{A r c} \textrm{\tan} \left(\frac{1}{- 3}\right) + {180}^{\circ}$

$- 3 + i = \setminus \sqrt{10} \setminus \setminus \textrm{c i s} \left(\textrm{A r c} \textrm{\tan} \left(\frac{1}{- 3}\right) + {180}^{\circ}\right)$

De Moivre doesn't work very well on a form like this, we get

${\left(- 3 + i\right)}^{3} = \setminus \sqrt{{10}^{3}} \setminus \setminus \textrm{c i s} \left(3 \left(\textrm{A r c} \textrm{\tan} \left(\frac{1}{- 3}\right) + {180}^{\circ}\right)\right)$

But we're not stuck. Since the exponent is only $3$ we can do this with triple angle formulas. Let's call the constant angle we found

$\theta = \angle \left(- 3 + i\right)$

By De Moivre,

${\left(- 3 + i\right)}^{3} = {\left(\setminus \sqrt{10} \setminus \setminus \textrm{c i s} \theta\right)}^{3} = 10 \sqrt{10} \left(\setminus \cos \left(3 \theta\right) + i \setminus \sin \left(3 \theta\right)\right)$

We know

$\setminus \cos \theta = - \frac{3}{\sqrt{10}} , \quad \sin \theta = \frac{1}{\sqrt{10}}$

$\cos \left(3 \theta\right) = 4 {\cos}^{3} \theta - 3 \cos \theta = 4 {\left(- \frac{3}{\sqrt{10}}\right)}^{3} - 3 \left(- \frac{3}{\sqrt{10}}\right) = - \frac{9 \sqrt{10}}{50}$

$\sin \left(3 \theta\right) = 3 \sin \theta - 4 {\sin}^{3} \theta = 3 \left(\frac{1}{\sqrt{10}}\right) - 4 {\left(\frac{1}{\sqrt{10}}\right)}^{3} = \frac{13 \sqrt{10}}{50}$

${\left(- 3 + i\right)}^{3} = 10 \sqrt{10} \left(\frac{\sqrt{10}}{50}\right) \left(- 9 + 13 i\right) = - 18 + 26 i$

That seems like much more work than just cubing $\left(- 3 + i\right) :$

 (-3+i)(-3+i)(-3+i)=(-3 + i)(8 -6i) = -18 + 26 i quad sqrt

OK, let's do the problem:

$\frac{{\left(2 + 2 i\right)}^{5} {\left(- 3 + i\right)}^{3}}{{\left(\sqrt{3} + i\right)}^{10}}$

$= \frac{{\left(2 \sqrt{2} \setminus \setminus \textrm{c i s} \setminus {45}^{\circ}\right)}^{5} {\left(- 3 + i\right)}^{3}}{{\left(2 \setminus \setminus \textrm{c i s} \setminus {30}^{\circ}\right)}^{10}}$

 = ( { 2^5 sqrt{2^5 }}/2^10 ) { \text{cis}(5 cdot 45^circ ) }/{\text{cis}(10 cdot 30^circ) } (-3 + i)^3

 = (sqrt{2}/8) { \text{cis}(225^circ ) }/{\text{cis}( 300^circ) } (-3 + i)^3

$= \left(\frac{\sqrt{2}}{8}\right) \setminus \textrm{c i s} \left({225}^{\circ} - 300\right) {\left(- 3 + i\right)}^{3}$

$= \left(\frac{\sqrt{2}}{8}\right) \left(- 18 + 26 i\right) \setminus \textrm{c i s} \left(- {75}^{\circ}\right)$

Ugh, it never ends. We get

$\cos \left(- {75}^{\circ}\right) = \cos {75}^{\circ} = \cos \left({45}^{\circ} + {30}^{\circ}\right) = \setminus \frac{\sqrt{2}}{2} \left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right) = \frac{1}{4} \left(\sqrt{6} - \sqrt{2}\right)$

$\sin \left(- {75}^{\circ}\right) = - \left(\sin 45 \cos 30 + \cos 45 \sin 30\right) = - \frac{\sqrt{2}}{2} \left(\setminus \frac{\sqrt{3}}{2} + \frac{1}{2}\right) = - \frac{1}{4} \left(\setminus \sqrt{6} + \sqrt{2}\right)$

$\frac{{\left(2 + 2 i\right)}^{5} {\left(- 3 + i\right)}^{3}}{{\left(\sqrt{3} + i\right)}^{10}}$

$= \left(\frac{\sqrt{2}}{8}\right) \left(- 18 + 26 i\right) \frac{1}{4} \left(\left(\sqrt{6} - \sqrt{2}\right) - \left(\sqrt{6} + \sqrt{2}\right) i\right)$

$= \frac{11 + 2 \sqrt{3}}{4} + \frac{11 \sqrt{3} - 2}{4} i$