# How do you divide ( i-3) / (5i +2) in trigonometric form?

Jan 5, 2017

$\frac{i - 3}{5 i + 2} \approx 0.587 \cdot \left(\cos \left(1.630\right) + i \cdot \sin \left(- 1.630\right)\right)$

#### Explanation:

$\textcolor{red}{\text{~~~ Complex Conversion: Rectangular " harr " Tigonometric~~~}}$
$\textcolor{w h i t e}{\text{XX }} \textcolor{b l u e}{a + b i \leftrightarrow r \cdot \left[\cos \left(\theta\right) + i \cdot \sin \left(\theta\right)\right]}$
$\textcolor{w h i t e}{\text{XXX")color(blue)("where } r = \sqrt{{a}^{2} + {b}^{2}}}$
color(white)("XXX")color(blue)("and "theta={("arctan"(b/a),"if "(a,bi) in "Q I or Q IV"),("arctan"(b/a)+pi,"if " (a,bi) in "Q II or Q III):})

$\textcolor{red}{\text{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}}$

$\textcolor{red}{\text{~~~ Trigonometric Division ~~~}}$
color(white)("XX ")color(blue)((cos(theta)+i * sin(theta))/(cos(phi)+i * sin(phi)))

color(white)("XXX")color(blue)(= [color(green)((cos(theta) * cos(phi) + sin(theta) * sin(phi))] +i * [color(magenta)(sin(theta) * cos(phi)-cos(theta) * sin(phi))]

$\textcolor{red}{\text{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}}$

If $A = i - 3 = - 3 + 1 i$
then
$\textcolor{w h i t e}{\text{XXX}} {r}_{A} = \sqrt{{\left(- 3\right)}^{2} + {1}^{2}} = \sqrt{10}$
and
$\textcolor{w h i t e}{\text{XXX}} {\theta}_{A} = \arctan \left(- \frac{1}{3}\right) + \pi \approx 2.819842099$

If $B = 5 i + 2 = 2 + 5 i$
then
$\textcolor{w h i t e}{\text{XXX}} {r}_{B} = \sqrt{{2}^{2} + {5}^{2}} = \sqrt{29}$
and
$\textcolor{w h i t e}{\text{XXX}} {\theta}_{B} = \arctan \left(\frac{5}{2}\right) \approx 1.19028995$

$\frac{A}{B} = \frac{i - 3}{5 + 2 i}$

$\textcolor{w h i t e}{\text{XXX}} = \frac{\sqrt{10}}{\sqrt{29}} \cdot \left(\frac{\cos \left({\theta}_{A}\right) + i \sin \left({\theta}_{A}\right)}{\cos \left({\theta}_{B}\right) + i \sin \left({\theta}_{B}\right)}\right)$

color(white)("XXX")=sqrt(10)/sqrt(29) *([color(green)(cos(theta_A)*cos(theta_B)+sin(theta_A) * sin(theta_B))+i[color(magenta)(sin(theta_A) * cos(theta_B) - cos(theta_A) * sin(theta_B))])

Plugging in the values for ${\theta}_{A}$ and ${\theta}_{B}$ and using a calculator/spreadsheet:
$\textcolor{w h i t e}{\text{XXX}} \approx - 0.034482759 + 0.586206897 i$

If we call this value $C$
then
$\textcolor{w h i t e}{\text{XXX}} {r}_{C} = \sqrt{{\left(- 0.03448 \ldots\right)}^{2} + {\left(0.5862 \ldots\right)}^{2}} \approx 0.58722022$
and (since $C$ is in Quadrant 2)
$\textcolor{w h i t e}{\text{XXX}} {\theta}_{C} = \arctan \left(\frac{0.5862 \ldots}{- 0.03448 \ldots}\right) + \pi \approx 1.62955215$

$\frac{A}{B} = C = {r}_{C} \left(\cos \left({\theta}_{C}\right) + i \cdot \sin \left({\theta}_{C}\right)\right)$

Jan 5, 2017

$\frac{i - 3}{5 i + 2} \approx 0.587 \left(\cos \left(1.630\right) + i \cdot \sin \left(1.630\right)\right)$

#### Explanation:

This is an alternate (and I think simpler solution) than the first one given (probably below this one) but it does not do the division in trigonometric form; instead, it does the division and then converts the result into trigonometric form.

$\frac{i - 3}{5 i + 2} = \frac{\left(i - 3\right) \cdot \left(5 i - 2\right)}{\left(5 i + 2\right) \cdot \left(5 i - 2\right)}$

color(white)("XXXXXXXXX"){:(underline(xx),underline("|"),underline(i),underline(-3)),(5i,"|",-5,-15i),(underline(-2),underline("|"),underline(-2i),underline(+6)),(,,1,-17i):} color(white)("XXXX"){:(underline(xx),underline("|"),underline(5i),underline(+2)),(5i,"|",-25,+10i),(underline(-2),underline("|"),underline(-10i),underline(-4)),(,,-29,):}

$\frac{i - 3}{5 i + 2} = \frac{1 - 17 i}{- 29} = - \frac{1}{29} + i \cdot \frac{17}{29}$

If we call this value $C$
then
$\textcolor{w h i t e}{\text{XXX}} {r}_{C} = \sqrt{{\left(\frac{1}{29}\right)}^{2} + {\left(\frac{17}{29}\right)}^{2}} \approx 0.58722022$
and (since $C$ is in Quadrant II)
$\textcolor{w h i t e}{\text{XXX}} {\theta}_{C} = \arctan \left(\frac{17}{- 1}\right) + \pi \approx 1.62955215$