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Question : 9 of 160
Marks:
+1,
-0
Solution:
‌z=‌| 1−i‌cos‌θ |
| 1+2isin‌θ |
‌ is purely imaginary ‌‌∴z+z=0‌⇒‌| 1−i‌cos‌θ |
| 1+2isin‌θ |
+‌| 1+i‌cos‌θ |
| 1−2isin‌θ |
=0‌1−2isin‌θ−i‌cos‌θ−2sin‌θ‌cos‌θ+1‌+i‌cos‌θ+2isin‌θ−2sin‌θ‌cos‌θ‌‌‌(1+2isin‌θ)(1−2isin‌θ)‌⇒(2−4sin‌θ‌cos‌θ)=0‌⇒2sin‌2θ=2⇒sin‌2θ=1‌⇒sin‌2θ=sin‌‌‌∴‌‌2θ=nπ+(−1)n‌,n∈z‌∴‌‌θ=‌+(−1)n‌,n∈z.
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