Show that the multiplication of any complex number, z, by e^jθ is describable, in geometric terms, as a positive rotation through the angle θ of the vector by which z is represented, without any alteration of its length.
Im not quite sure what to do. I do have the equations
e^jθ =cosθ +jsinθ and z=a+jb
thanks in advance!
Im not quite sure what to do. I do have the equations
e^jθ =cosθ +jsinθ and z=a+jb
thanks in advance!
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e^jθ =cosθ +jsinθ and z=a+jb
The magnitude of e^jθ = 1
|e^jθ|= √((cos θ)^2 +((sin θ)^2) = 1
If you multiply z(e^jθ) = (a +jb)(cosθ +jsinθ) you get
z(e^jθ) = a(cosθ) +a (jsin θ) -bsin θ +b(jcos θ)
The new A = acosθ - bsin θ and the new B =j(b cosθ +asin (θ))
The magnitude of new Z = √(A^2 + B^2) = √ ((a cosθ - b sin θ)^2 + ( b cos θ + a sin θ) ^2)
Z = √ ((a^2 cos^2 θ + a^2 sin^2 θ + b^2 cos^2 θ + b^2 sin^2 θ)
since cos^2 θ + sin^2 θ =1
Z = √(a^2 +b^2)
so the new |Z| = |z|
The magnitude of e^jθ = 1
|e^jθ|= √((cos θ)^2 +((sin θ)^2) = 1
If you multiply z(e^jθ) = (a +jb)(cosθ +jsinθ) you get
z(e^jθ) = a(cosθ) +a (jsin θ) -bsin θ +b(jcos θ)
The new A = acosθ - bsin θ and the new B =j(b cosθ +asin (θ))
The magnitude of new Z = √(A^2 + B^2) = √ ((a cosθ - b sin θ)^2 + ( b cos θ + a sin θ) ^2)
Z = √ ((a^2 cos^2 θ + a^2 sin^2 θ + b^2 cos^2 θ + b^2 sin^2 θ)
since cos^2 θ + sin^2 θ =1
Z = √(a^2 +b^2)
so the new |Z| = |z|