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Question : 9 of 160
Marks:
+1,
-0
Solution:
z−4=(8i)1∕3 z=4+2(i)1∕3 i=eiπ∕2=ei(‌+2nπ) i1∕3=ei(‌+‌),n=0,1,2 [| ei‌=cos‌+i‌sin‌ | =‌+i‌ |
| ei(‌+‌) | =ei‌=cos‌+i‌sin‌=‌+i‌ |
| ei(‌+‌) | =ei‌ |
| ‌ | =cos‌+i‌sin‌=0−i |
]} For,
n=0,eiπ∕6=‌+i‌ n‌‌=1,ei5π∕6=−‌+i⋅‌ n‌‌=2,ei‌=0−i ∴‌‌z‌‌=4+2(‌+i‌)=(4+√3)+i=b+i z‌‌=4+2(‌+‌)=(4−√3)+i=c+i ⇒‌‌z‌‌=4+2(0−i)=4−2i=a−2i ∴‌‌a‌‌=4,b=4+√3‌ and ‌c=4−√3 ∴‌‌4(4+√3)(4−√3)‌‌=√4(16−3)=2√13
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