Question 896411
• Let x = r(cos u + i sin u)
• Let y = t(cos v + i sin v)
Prove that xy = rt(cos(u+v) + i sin(u+v)) and prove that
The radius (or modulus) of the product xy is rt
I have to solve and justify the steps. I know that I will plug in the cos u.... and cos v... I have read the chapter and stared at the work for awhile but I am not getting anywhere!


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x = r(cos u + i sin u)
y = t(cos v + i sin v)




x*y = r(cos u + i sin u)*t(cos v + i sin v)


x*y = r*t(cos u + i sin u)*(cos v + i sin v)


x*y = r*t[ (cos u + i sin u)*(cos v + i sin v) ]


x*y = r*t[ cos(u)*(cos(v) + i*sin(v))+i*sin(u)*(cos(v) + i*sin(v)) ]


x*y = r*t[ cos(u)*cos(v) + i*cos(u)*sin(v)+i*sin(u)*cos(v) + i*i*sin(u)*sin(v) ]


x*y = r*t[ cos(u)*cos(v) + i*cos(u)*sin(v)+i*sin(u)*cos(v) + i^2*sin(u)*sin(v) ]


x*y = r*t[ cos(u)*cos(v) + i*cos(u)*sin(v)+i*sin(u)*cos(v) + (-1)*sin(u)*sin(v) ]


x*y = r*t[ cos(u)*cos(v) + i*cos(u)*sin(v)+i*sin(u)*cos(v) - sin(u)*sin(v) ]


x*y = r*t[ cos(u)*cos(v) - sin(u)*sin(v) + i*cos(u)*sin(v)+i*sin(u)*cos(v) ]


x*y = r*t[ (cos(u)*cos(v) - sin(u)*sin(v)) + (i*cos(u)*sin(v)+i*sin(u)*cos(v)) ]


x*y = r*t[ cos(u+v) + i*(cos(u)*sin(v)+sin(u)*cos(v)) ]


x*y = r*t[ cos(u+v) + i*sin(u+v) ]



This number x*y is in the form r(cos(theta) + i*sin(theta)) where the radius is r*t and the angle is u+v