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If \(\rm iz^3+z^2-z+i=0\)then |z| equal to 1. 12. 23. -24. √2 |
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Answer» Correct Answer - Option 1 : 1 Concept: Properties of iota: i2 = -1 \(\rm \frac 1 i = \frac i {i^2} = \frac i {-1} = -i\)
Calculation: We have, \(\rm iz^3+z^2-z+i=0\) On dividing by i, we get \(⇒ \rm z^3+\frac {z^2} {i}-\frac z i+1=0\) \(⇒ \rm z^3-iz^2+iz+1=0\) (∵ \(\rm \frac 1 i \)= -i) \(⇒ \rm z^3-iz^2+iz-i^2=0\) (∵ i2 = -1) \(⇒ \rm z^2(z-i)+i(z-i)=0\) \(⇒ \rm (z^2+i)(z-i)=0\) So, z = i or z2 = -i. Now, z = i ⇒ |z| = |i| ⇒ |z| = 1 And, z2 = -i ⇒ |z2| = |-i| = 1 ⇒ |z| = 1 Hence, option (1) is correct. |
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