If z -sin- i cos- then for any integer IA- z n -1- Z n -2 cos n -2- n -B - z n -1- z n -2 sin n -2- n -C - z n -1- Z n -2 i sin n - n -2D - z n -1- Z n -2 i cos n - n -2

The correct options are A z^n+1z^n=2cos(nπ2 nθ) C z^n 1z^n=2isin(nθ nπ2)Here, we can write the complex number z in Euler's form as z=sinθ icosθ z=cos(θ π2)+isin ...

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Standard XII

Mathematics

Algebra of Complex Numbers

If z =sinθ- i...

Question

If $z = sin θ - i cos θ$ , then for any integer $n$

z^{n} + \frac{1}{z^{n}} = 2 cos (\frac{n π}{2} - n θ)

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z^{n} + \frac{1}{z^{n}} = 2 sin (\frac{n π}{2} - n θ)

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z^{n} - \frac{1}{z^{n}} = 2 i sin (n θ - \frac{n π}{2})

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z^{n} - \frac{1}{z^{n}} = 2 i cos (n θ - \frac{n π}{2})

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Solution

The correct options are
A $z^{n} + \frac{1}{z^{n}} = 2 cos (\frac{n π}{2} - n θ)$
C $z^{n} - \frac{1}{z^{n}} = 2 i sin (n θ - \frac{n π}{2})$
Here, we can write the complex number $z$ in Euler's form as
$z = sin θ - i cos θ$
$z = cos (θ - \frac{π}{2}) + i sin (θ - \frac{π}{2})$
$\Rightarrow z = e^{i (θ - \frac{π}{2})}$
$\Rightarrow z^{n} = e^{i (n θ - \frac{n π}{2})}$
Similarly $z^{- n} = e^{i (- n θ + \frac{n π}{2})}$
Thus, adding both we get
$z^{n} + \frac{1}{z^{n}} = e^{i (n θ - \frac{n π}{2})} + e^{i (- n θ + \frac{n π}{2})}$
$z^{n} + \frac{1}{z^{n}} = cos (n θ - \frac{n π}{2}) + i sin (n θ - \frac{n π}{2}) + cos (n θ - \frac{n π}{2}) - i sin (n θ - \frac{n π}{2})$
$\Rightarrow z^{n} + \frac{1}{z^{n}} = 2 cos (\frac{n π}{2} - n θ)$
Similarly subtracting both we get
$z^{n} - \frac{1}{z^{n}} = 2 i sin (n θ - \frac{n π}{2})$

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