Correct Answer - Option 1 : 1 only
Concept:
Consider z = a + ib
Multiplicative identity z.z-1 = 1
zz̅ = a2 + b2 = (a + ib) (a - ib)
Explanation
Given:
1. \(\rm {\overline{\left(z^{-1}\right)}}=(\bar{z})^{-1}\)
consider z = a - ib
⇒ z-1 = \(\rm \frac 1 z = \frac 1 {a - ib}\) =\(\rm \frac{a + ib}{a^{2} + b^{2}}\)
⇒ \(\rm \overline{z^{-1}}\) = \(\rm \frac{a - ib}{a^{2} + b^{2}}\) ....(1)
⇒ \(\rm \overline{z}\) = a + ib
⇒ (\(\rm \overline{z}\))-1 = \(\rm \frac 1 {\bar{z}} = \frac 1 {a +ib}\)= \(\rm \frac{a - ib}{a^{2} + b^{2}}\) ....(2)
from eq(1) and (2)
statement 1 is correct.
2. zz-1 = |z|2
consider z = a - ib
⇒ z-1 = \(\rm \frac{a + ib}{a^{2} + b^{2}}\)
⇒ z.z-1 = (a - ib)\(\rm \frac{a + ib}{a^{2} + b^{2}}\) = \(\rm \frac{a^{2} + b^{2}}{a^{2} + b^{2}} =1\)
Statement 2 is not correct