Asked by Dani
If z=costheta +isintheta, prove that
a) z = 1/z
b) z+1/z = 2costheta
c) z-1/z=2isintheta
Hence ,deduce that z^2-1/z^2 +1=itantheta
a) z = 1/z
b) z+1/z = 2costheta
c) z-1/z=2isintheta
Hence ,deduce that z^2-1/z^2 +1=itantheta
Answers
Answered by
Reiny
z = cosθ + isinθ , then
a)
1/z = 1/(cosθ + isinθ) = 1/(cosθ + isinθ) * (cosθ - isinθ)/(cosθ - isinθ)
= (cosθ - isinθ)/(cos^2 θ - i^2 sin^2 θ)
= (cosθ - isinθ)/1 = <b>cosθ - isinθ</b>
b) z + 1/z = cosθ + isinθ + cosθ - isinθ = 2cosθ
c) z - 1/z = (cosθ + isinθ) - (cosθ - isinθ) = 2i sinθ
d) (z-1/z) / (z+1/z)
= ( (z^2 - 1)/z ) / ( (z^2 + 1)/z )
= ( (z^2 - 1)/z )*( z/(z^2 + 1) )
= (z^2 - 1)/(z^2 + 1) = 2isinθ/(2cosθ) = i tanθ
btw, did you notice that z ≠ 1/z ??
but rather, a), b), and c) are true for my results.
a)
1/z = 1/(cosθ + isinθ) = 1/(cosθ + isinθ) * (cosθ - isinθ)/(cosθ - isinθ)
= (cosθ - isinθ)/(cos^2 θ - i^2 sin^2 θ)
= (cosθ - isinθ)/1 = <b>cosθ - isinθ</b>
b) z + 1/z = cosθ + isinθ + cosθ - isinθ = 2cosθ
c) z - 1/z = (cosθ + isinθ) - (cosθ - isinθ) = 2i sinθ
d) (z-1/z) / (z+1/z)
= ( (z^2 - 1)/z ) / ( (z^2 + 1)/z )
= ( (z^2 - 1)/z )*( z/(z^2 + 1) )
= (z^2 - 1)/(z^2 + 1) = 2isinθ/(2cosθ) = i tanθ
btw, did you notice that z ≠ 1/z ??
but rather, a), b), and c) are true for my results.
Answered by
oobleck
I think he had trouble with z* or something for z-conjugate.
overbars are hard to get right.
overbars are hard to get right.
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