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Question : 7 of 160
Marks:
+1,
-0
Solution:
iz3+z2−z+i=0 ⇒iz3−z+z2+i=0⇒iz3+i2z+z2+i=0⇒iz(z2+i)+(z2+i)=0⇒(z2+i)(iz+1)=0⇒(z2+i)(iz−i2)=0⇒(z2+i)i(z−i)=0⇒z2=−i or z=iz2=1(cos+isin) or z=1(cos+isin) We can see that all the roots have modulus as
So
|7|=1
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