InterviewSolution
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InterviewSolution
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If `(3pi)/(2) lt alpha lt 2 pi`, find the modulus and argument of `(1 - cos 2 alpha) + i sin 2 alpha `. |
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Answer» `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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