The moduli of the numerator and denominator are |3i-7|=sqrt(3^2+(-7)^2)=sqrt(9+49)=sqrt(58)|3i−7|=√32+(−7)2=√9+49=√58 and
|2i-1|=sqrt(2^2+(-1)^2)=sqrt(5)|2i−1|=√22+(−1)2=√5. Therefore, the modulus of (3i-7)/(2i-1)3i−72i−1 is sqrt(58)/sqrt(5)=(sqrt(58)sqrt(5))/5=sqrt(290)/5√58√5=√58√55=√2905.
Both the numerator and the denominator sit in the 2nd quadrant of the complex plane. We can use the inverse (arc) cosine function to find their arguments (polar angles) most directly (we could also use the inverse (arc) tangent function but we'd have to add piπ to its output).
The argument of 3i-7=-7+3i3i−7=−7+3i is cos^{-1}(-7/sqrt(58))cos−1(−7√58) and the argument of 2i-1=-1+2i2i−1=−1+2i is cos^{-1}(-1/sqrt(5))cos−1(−1√5). Therefore, the argument of (3i-7)/(2i-1)3i−72i−1 is cos^{-1}(-7/sqrt(58))-cos^{-1}(-1/sqrt(5))cos−1(−7√58)−cos−1(−1√5).
Now cos(cos^{-1}(-7/sqrt(58)))=-7/sqrt(58)cos(cos−1(−7√58))=−7√58, sin(cos^{-1}(-7/sqrt(58)))=sqrt(1-(-7/sqrt(58))^2)=sqrt(9/58)=3/sqrt(58)sin(cos−1(−7√58))=√1−(−7√58)2=√958=3√58,
cos(cos^{-1}(-1/sqrt(5)))=-1/sqrt(5)cos(cos−1(−1√5))=−1√5, and
sin(cos^{-1}(-1/sqrt(5)))=sqrt(1-(-1/sqrt(5))^2)=sqrt{4/5}=2/sqrt(5)sin(cos−1(−1√5))=√1−(−1√5)2=√45=2√5.
Furthermore, cos(alpha-beta)=cos(alpha)cos(beta)+sin(alpha)sin(beta)cos(α−β)=cos(α)cos(β)+sin(α)sin(β), which implies that
cos(cos^{-1}(-7/sqrt(58))-cos^{-1}(-1/sqrt(5)))=
cos(cos^{-1}(-7/sqrt(58)))cos(cos^{-1}(-1/sqrt(5)))+sin(cos^{-1}(-7/sqrt(58)))sin(cos^{-1}(-1/sqrt(5)))cos(cos−1(−7√58)−cos−1(−1√5))=cos(cos−1(−7√58))cos(cos−1(−1√5))+sin(cos−1(−7√58))sin(cos−1(−1√5))
=-7/sqrt(58) * (-1/sqrt(5))+3/sqrt(58) * 2/sqrt(5)=13/sqrt(290)=−7√58⋅(−1√5)+3√58⋅2√5=13√290.
In a similar way, sin(alpha-beta)=sin(alpha)cos(beta)-cos(alpha)sin(beta)sin(α−β)=sin(α)cos(β)−cos(α)sin(β), which implies that
sin(cos^{-1}(-7/sqrt(58))-cos^{-1}(-1/sqrt(5)))=
sin(cos^{-1}(-7/sqrt(58)))cos(cos^{-1}(-1/sqrt(5)))-cos(cos^{-1}(-7/sqrt(58)))sin(cos^{-1}(-1/sqrt(5)))sin(cos−1(−7√58)−cos−1(−1√5))=sin(cos−1(−7√58))cos(cos−1(−1√5))−cos(cos−1(−7√58))sin(cos−1(−1√5))
=3/sqrt(58) * (-1/sqrt(5))-(-7)/sqrt(58) * 2/sqrt(5)=11/sqrt(290)=3√58⋅(−1√5)−−7√58⋅2√5=11√290.
Finally, Euler's formula allows us to put all this together to say that
(3i-7)/(2i-1)=|z|cos(theta)+i |z|sin(theta)3i−72i−1=|z|cos(θ)+i|z|sin(θ)
=sqrt(290)/5 * 13/sqrt(290)+i*sqrt(290)/5 * 11/sqrt(290)=13/5+11/5 i=√2905⋅13√290+i⋅√2905⋅11√290=135+115i.
