A) 0
B) 1
C) 2
D) None of these
Correct Answer: B
Solution :
[b]: Given, \[i{{z}^{3}}+{{z}^{2}}-z+i=0\] Dividing both side by f and using \[\frac{1}{i}=-i\], we have \[{{z}^{3}}-i{{z}^{2}}+iz+1=0\] \[\Rightarrow \]\[{{z}^{2}}(z-i)+i(z-i)=0\] \[(\because {{i}^{2}}=-1)\] \[\Rightarrow \]\[(z-i)({{z}^{2}}+i)=0\therefore z=i\]or\[{{z}^{2}}=-i\] \[\therefore \]\[|z|=|i|=1\]and\[|{{z}^{2}}|=|z{{|}^{2}}=|-1|=1\therefore |z|=1\]
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