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If (3pi)/2 < alpha < 2pi then the modulu...

If `(3pi)/2` < `alpha` < `2pi` then the modulus argument of `(1+cos 2alpha)+i sin2alpha`

Text Solution

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`z = (1-cos2 alpha)+ i sin 2alpha`
`= 2sin^(2) alpha + i (2sin alpha cos alpha)`
`=2 sin alpha(sin alpha + i cos alpha)`
`=-2 sin alpha (-sin alpha - i cos alpha)`
`=-2 sin alpha (cos((3pi)/(2)-alpha)+ sin ((3pi)/(2)-alpha))`
Thus, `|z| = - s sin alpha`
Alos, `(3pi)/(2) - alpha in ((-pi)/(2),0)`
Therefore, z lies in the fourth quadrant.
`therefore arg(z) = (3pi)/(2) =- alpha`
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