1.

If \(\rm iz^3+z^2-z+i=0\)then |z| equal to 1. 12. 23. -24. √2

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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