Precalculus Question

I have another method :)

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

where theta is in radians

$$\\148^0=\frac{148\pi}{180}\; radians\\\\ 11^0=\frac{11\pi}{180}\; radians\\\\$$

Given Z = 2(cos 148º + isin 148º) and W = 5(cos 11º + isin 11º), find and simplify Z divided by W.

becomes  

$$\\\dfrac{2e^{(148\pi/180)i}}{5e^{(11\pi/180)i}}\\\\ =0.4e^{[(148\pi/180)-(11\pi/180)]i}\\\\$$

$$\left({\frac{{\mathtt{148}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{180}}}}\right){\mathtt{\,-\,}}\left({\frac{{\mathtt{11}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{180}}}}\right) = {\mathtt{2.391\: \!101\: \!075\: \!232\: \!231\: \!5}}$$

=0.4cos (2.3911010752322315) +0.4* isin(2.3911010752322315) 

remember this is in radians.

= -0.2925 + 0.2728i

=  -0.29 + 0.27i

I think that method is correct. 

If you have two complex numbers written in polar notation, for example:

A = a(cosα + i·sinα)     and     B = b(cosβ + i·sinβ)  

Then A / B  =  (a/b)( cos(α - β) + i·sin(α - β) )

Can you see how to apply this?

How did you get your equation Geno.

I can see another method but you seem to have used a 3rd method 

To answer your question Melody, use the following notation;

$$A=ae^{i\alpha}$$

$$B=be^{i\beta}$$

then

$$\frac{A}{B}=\frac{a}{b}e^{i(\alpha - \beta)}\rightarrow \frac{a}{b}(\cos(\alpha -\beta)+i\sin(\alpha - \beta))$$

.

There you go.

I am practicing my philosophical approach.

Why do it the easy way if there is a long way that works just as well.    LOL

Thanks Alan.