How do you prove: z/w = r/R[  (cos (θ - Θ) + i sin (θ - Θ))]  

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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Write the polar form of two complex numbers z and w.

`z = r(cos theta + i sin theta)`

`w = R(cosTheta + i sin Theta)`

Divide z by w:

`z/w =(r(cos theta + i sin theta))/(R(cosTheta + i sin Theta))`

`` Multiply both the denominator and numerator by the conjugate of denominator `R(cosTheta- i sin Theta).`

`z/w =(r(cos theta + i sin theta)(cosTheta- i sin Theta))/(R(cosTheta + i sin Theta)(cosTheta- i sin Theta))`

Removing the brackets yields:

`z/w =r(cos theta*cos Theta- i*cos theta*sin Theta +i sin theta*cos Theta - i^2 sin theta* sin Theta)/(R(cos^2Theta- i^2 sin^2 Theta))`

Use complex number theory and replace `i^2`  by -1.

`z/w =r(cos theta*cos Theta- i*cos theta*sin Theta +i sin theta*cos Theta+ sin theta* sin Theta)/(R(cos^2Theta + sin^2 Theta))`

Use basic formula of trigonometry `cos^2Theta + sin^2 Theta = 1` .

`z/w =(r/R)*(cos theta*cos Theta + sin theta* sin Theta- i*(cos theta*sin Theta-sin theta*cos Theta)`

`cos theta*cos Theta + sin theta* sin Theta = cos(theta - Theta)`

`cos theta*sin Theta-sin theta*cos Theta= sin (Theta - theta) = - sin (theta - Theta)`

`z/w =(r/R)*(cos(theta - Theta) - i*(- sin (theta - Theta)))`

`z/w =(r/R)*(cos(theta - Theta)+ i*sin (theta - Theta))`

The last line proves that `z/w =(r/R)*(cos(theta - Theta)+ i*sin (theta - Theta))`

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