the principal amplitude of (sin 40°+icos40°)^5 is
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Given , z=(sin40+icos40)5⇒z = (cos 50 + i sin 50)5⇒z = cos 250 + i sin 250⇒z = cos [360 − 110] + i sin [360−110]⇒z = cos 110 − i sin 110⇒z = cos (−110) + i sin (−110)So, amp z = −110°
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