heya gyzz.plzz solve this ques of class 11th frm ch :-complex nos
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convert it into cos x + isin x form then use demoiver theorm for both brackets diff...u can find ur ans by dos way i guess...thanx for reading mam
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Answered by
6
answer:
(1+i)6=((1+i)2)3(1+i)6=((1+i)2)3
=(12+i2+2i)3=(12+i2+2i)3
=(1−1+2i)3=8i3=−8i=(1−1+2i)3=8i3=−8i
(1−i)3=13−i3−3i+3i2(1−i)3=13−i3−3i+3i2
=1+i−3i−3=−2−2i=1+i−3i−3=−2−2i
(1+i)6+(1−i)3=−8i−2−2i(1+i)6+(1−i)3=−8i−2−2i
=−2−10i.
(1+i)6=((1+i)2)3(1+i)6=((1+i)2)3
=(12+i2+2i)3=(12+i2+2i)3
=(1−1+2i)3=8i3=−8i=(1−1+2i)3=8i3=−8i
(1−i)3=13−i3−3i+3i2(1−i)3=13−i3−3i+3i2
=1+i−3i−3=−2−2i=1+i−3i−3=−2−2i
(1+i)6+(1−i)3=−8i−2−2i(1+i)6+(1−i)3=−8i−2−2i
=−2−10i.
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Answered by
0
Answer:
answer:
(1+i)6=((1+i)2)3(1+i)6=((1+i)2)3
=(12+i2+2i)3=(12+i2+2i)3
=(1−1+2i)3=8i3=−8i=(1−1+2i)3=8i3=−8i
(1−i)3=13−i3−3i+3i2(1−i)3=13−i3−3i+3i2
=1+i−3i−3=−2−2i=1+i−3i−3=−2−2i
(1+i)6+(1−i)3=−8i−2−2i(1+i)6+(1−i)3=−8i−2−2i
=−2−10i.
Hope the answer is help ful for you
Please brainlest me the answer
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