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Question : 9 of 160
Marks:
+1,
-0
Solution:
Let,
Z=√√2+1+i√√2−1‌⇒‌‌|Z|=√√2+1+√2−1=√2√2‌‌‌cos‌θ=‌,sin‌θ=‌‌∴‌‌cos2θ=‌,sin‌2θ=‌‌⇒‌=‌‌⇒cos‌2‌θ=‌⇒2θ=‌ (since, real and imaginary part are positive.)
Now,
‌Z8=(√2√2)8(cos(8×‌)+isin‌(8×‌))‌=(2√2)4(cos‌π+isin‌π)‌=(2‌×4)(−1+0)‌=26(−1)=−64
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