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Question : 69 of 160
Marks:
+1,
-0
Solution:
‌z=1+i√3 ‌‌ Let ‌1+i√3=r(cos‌θ+isin‌θ) ‌‌ Then, ‌r‌cos‌θ=1 . . . (i)
‌r‌sin‌θ=√3 . . . (ii)
Eq. (i) + Eq. (ii), we get
‌r2(sin‌2θ+cos2θ)=3+1(∵sin‌2θ+cos2θ=1)‌r2=4 r=2Eq. (ii)
÷ Eq. (i), we get
⇒tan‌θ=√3⇒tan‌θ=tan‌⇒θ=‌So,
z=(1+i√3)=2{cos‌+isin‌‌},
Then
z=(1−i√3)=2{cos‌−isin‌‌}=2{cos(‌)+isin‌(‌)}So ‌Arg‌z=‌‌Arg‌z=−‌|Arg‌z|+|Arg‌z|‌=|‌|+|‌|‌=‌+‌ =‌
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