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$${\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{i}$$

Guest Nov 17, 2014

Best Answer 

 #3
avatar+26625 
+10

Note that heureka's answer and mine are different ways of expressing the same thing because

$$e^{i\theta}=cos\theta+isin\theta$$

.

Alan  Nov 17, 2014
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3+0 Answers

 #1
avatar+26625 
+10

rectangular form: x + iy

polar form: re

r = √(x2 + y2)

θ = tan-1(y/x)

 

$${\mathtt{r}} = {\sqrt{{{\mathtt{2}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{3}}}^{{\mathtt{2}}}}} \Rightarrow {\mathtt{r}} = {\mathtt{3.605\: \!551\: \!275\: \!463\: \!989\: \!3}}$$

$${\mathtt{theta}} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{3}}}{{\mathtt{2}}}}\right)} \Rightarrow {\mathtt{theta}} = {\mathtt{56.309\: \!932\: \!474\: \!02^{\circ}}}$$

.

Alan  Nov 17, 2014
 #2
avatar+19206 
+8

how to conert rectangle form 2+3i to polar form :

$$z = x + y i \quad | \quad z = 2 + 3i$$

polar: $$x = r*\cos{(\phi)} \quad and \quad y = r*\sin{(\phi)}$$

set into z = x + yi:  $$z = r*\cos{(\phi)}+ r*\sin{(\phi)}i = r \left[ \cos{(\phi)}+ \sin{(\phi)}i \right] = r \left[ \cos{(\phi)}+ i\sin{(\phi)} \right]$$

r:  $$r=\sqrt{x^2+y^2} \quad | \quad r = \sqrt{2^2+3^2} = \sqrt{4+9} = \sqrt{13}$$ 

$$\phi$$ : $$\phi = \tan^{-1}{ ( \frac{y}{x} )} \quad | \quad \phi = \tan^{-1}(\frac{3}{2}) = 56.3099324740\ensurement{^{\circ}}$$

z(polar): $$z=r \left[ \cos{(\phi)}+ i\sin{(\phi)} \right] \quad | \quad z = \sqrt{13} \left[ \cos{(56.3099324740\ensurement{^{\circ}})}+ i*\sin{(56.3099324740\ensurement{^{\circ}})} \right]$$

heureka  Nov 17, 2014
 #3
avatar+26625 
+10
Best Answer

Note that heureka's answer and mine are different ways of expressing the same thing because

$$e^{i\theta}=cos\theta+isin\theta$$

.

Alan  Nov 17, 2014

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