@CPhill

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.

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

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)

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

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

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   

Thanks Tiggsy,

It is always great to see you on the forum