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

Jan 4, 2017

$\frac{- i - 2}{2 i - 5} \textcolor{g r e e n}{\approx 0.27586 - i \cdot 0.310345}$

#### Explanation:

Information that will be useful:
$\textcolor{red}{\text{~~~ Complex Conversion: Rectangular " harr " Tigonometric~~~}}$
$\textcolor{w h i t e}{\text{XX }} \textcolor{b l u e}{a + b i \leftrightarrow r \cdot \left[\cos \left(\theta\right) + i \cdot \sin \left(\theta\right)\right]}$
$\textcolor{w h i t e}{\text{XXX")color(blue)("where } r = \sqrt{{a}^{2} + {b}^{2}}}$
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))]

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

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If $A = - i - 2$
color(white)("XXX")r_A = sqrt((-1)^2+(-2)^) = sqrt(5)
and (since $\left(- 1 , - 2\right)$ is in Quadrant III)
color(white)("XXX")theta_A = "arctan"((-2)/(-1))+pi = arctan(2)+pi
$\textcolor{w h i t e}{\text{XXXXX}} \approx 4.248741371$ (with the aid of a caculator)

If $B = 2 i - 5$
$\textcolor{w h i t e}{\text{XXX}} {r}_{B} = \sqrt{{2}^{2} + {\left(- 5\right)}^{2}} = \sqrt{29}$
and
color(white)("XXX")theta_B="arctan"(-5/2)
$\textcolor{w h i t e}{\text{XXXXX}} \approx - 1.19028995$ (again with calculator)

So $\frac{A}{B} = \frac{- i - 2}{2 i - 5} = \frac{\sqrt{5} \cdot \left[\cos \left(4.248 \ldots\right) + i \cdot \sin \left(4.248 \ldots\right)\right]}{\sqrt{29} \cdot \left[\cos \left(- 1.190 \ldots\right) + i \cdot \sin \left(- 1.190 \ldots\right)\right]}$

$\textcolor{w h i t e}{\text{XXX}} \approx \left[\cos \left(4.249\right) \cdot \cos \left(- 1.190\right) + \sin \left(4.249\right) \cdot \sin \left(- 1.190\right)\right] + i \left[\sin \left(4.248\right) \cdot \cos \left(1.190\right) - \cos \left(4.248\right) \cdot \sin \left(- 1.190\right)\right]$

(and, once more using a calculator)
$\textcolor{w h i t e}{\text{XXX}} \approx 0.2758620689 - i \left(0.3103448276\right)$