SOLUTION: • 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 ha

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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!

Found 2 solutions by jim_thompson5910, Edwin McCravy:
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
• 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

Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!



Multiply equals by equals:





Use FOIL



since 





factor i out of the middle two terms



Move the last term next to the first term:



Use double angle formulas 
and 

You might swap the terms of the imaginary part:



and you have:



Edwin

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