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If you do this could you please put and explain every step and how you did it??? Thanks!!!

 

find each quotient and write it in rectangular form

 

\(\frac{12cis293}{6cis23}\)

 Dec 21, 2017
 #1
avatar+109520 
+1

 

I have asssumed that the angles are in degrees ...which i really should not have done :/

I expect there is a much easier way to do this. Also the question implies there should be more than one answer. My knowledge here is sketchy at best.

 

\(\frac{12cis293}{6cis23}\\ \qquad cis293\\ \qquad=cis(270+23)\\ \qquad=cos(270+23)+isin(270+23)\\ \qquad=cos(270)cos(23)-sin(270)sin(23)+i[sin(270)cos(23)+cos(270)sin(23)]\\ \qquad=0*cos(23)--1sin(23)+i[-1*cos(23)+0*sin(23)]\\ \qquad=sin(23)+i[-cos(23)]\\ \qquad=sin(23)-i[cos(23)]\\~\\ \frac{12cis293}{6cis23}\\ =\frac{2[sin23-icos23]}{cos23+isin23}\\ =\frac{2[sin23-icos23]}{cos23+isin23}\times \frac{cos23-isin23}{cos23-isin23}\\ =\frac{2[sin23(cos23-isin23) -icos23(cos23-isin23) ]}{cos^223-i^2sin^223}\\ =\frac{2[sin23cos23-isin^223-icos^223+i^2sin23cos23 ]}{cos^223-i^2sin^223}\\ =\frac{2[sin23cos23-isin^223-icos^223-sin23cos23]}{cos^223+sin^223}\\ =\frac{2[sin23cos23-sin23cos23 -isin^223-icos^223]}{1}\\ =2[-i(sin^223+cos^223)\\ =-2i \)

 


 

 Dec 21, 2017
edited by Melody  Dec 21, 2017
 #2
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+2

Melody: I don't pretend to understand this, but "Mathematica 11" gives this answer without any steps or explanation. Sorry about that:

 

2 (cos(293) + i sin(293)))/(cos(23) + i sin(23))

=2 (cos(270) + i sin(270)), OR

=2 cos(270) + 2 i sin(270)

=2 e^(270 i)

 Dec 21, 2017
 #3
avatar+109520 
+2

Yes 

If the angles are in degrees then this becomes

2cos270=2*0=0

2isin270 =2i*-1 = -2i

 

2cos270+isin270=-2i      (Which is what I got)

 

But Mathematica's answer, the unsimplified answer,  is much better because the angle should be in radians rather than degrees. 

 

As far as there being no explanation given by Mathematica, well I am sure there are acceptable short cuts here. My knowledge in this area is limited.

Melody  Dec 21, 2017
edited by Melody  Dec 21, 2017
 #4
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+3

\(\displaystyle \text{If }z_{1}=r_{1}(\cos\theta_{1}+\imath\sin\theta_{1})\text{ and }z_{2}=r_{2}(\cos\theta_{2}+\imath\sin\theta_{2})\),

then

\(\displaystyle z_{1}z_{2}=r_{1}r_{2}\{\cos(\theta_{1}+\theta_{2})+\imath\sin(\theta_{1}+\theta_{2})\}\)

and

\(\displaystyle z_{1}/z_{2}=(r_{1}/r_{2})\{\cos(\theta_{1}-\theta_{2})+\imath\sin(\theta_{1}-\theta_{2})\}\).

 

To multiply, multiply the moduli and add the arguments, to divide, divide the moduli and subract one argument from the other.

They're both easy to prove using well-known trig identities.

 

If the numbers are written in exponential form via the identity

\(\displaystyle e^{\imath\theta}=\cos\theta+\imath\sin\theta\),

then the rules can be seen to become, (or), the rules for indices can be used to produce

\(\displaystyle z_{1}z_{2}=r_{1}r_{2}e^{\imath(\theta_{1}+\theta_{2})}\text{ and }z_{1}/z_{2}=(r_{1}/r_{2})e^{\imath(\theta_{1}-\theta_{2})}\).

 

Applying to the given problem, (cis is just an abbreviation for cos + i.sin),

 

\(\displaystyle 12\text{cis}293/6\text{cis}23=2\text{cis}270, \text{ (or, alternatively)}\text{ }2e^{\imath270}\).

 

Mathematica has not been told what the units are so assumes radians, personally though I take the view that the angles are in degrees, (270 degrees or 270 radians ?), in which case the result can be written as

\(\displaystyle 2(\cos270+\sin270)=2(0+\imath(-1))=-2\imath\).

 

Tiggsy

 Dec 21, 2017
 #5
avatar+109520 
+3

ok I have done some homework.

 

\(cis(A+B)=cis(A)\times cis(B)\qquad \text{From Euler's formula}\\ so\\ \frac{12cis293}{6cis23}=\frac{2cis270cis23}{cis23}=2cis270\\ =2cos270+2isin270\\ =2e^{(270i)} \)

 

 

Now I just want to prove this identity.

 

\(cis(A+B)\\=cos(A+B)+isin(A+B)\\ =cosAcosB-sinAsinB+isinAcosB+icosAsinB\\ =cosAcosB+icosAsinB+isinAcosB-sinAsinB\\ =cosAcosB+icosAsinB+isinAcosB+iisinAsinB\\ =cosA(cosB+isinB)+isinA(cosB+isinB)\\ =(cosA+isinA)(cosB+isinB)\\ =cisA\times cisB\\ QED \)

 

Now i am happy   laugh

 Dec 21, 2017
 #6
avatar+109520 
+1

Thanks Tiggsy,

 

It is always great to see you on the forum  laughlaughlaugh

 Dec 21, 2017

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