Question-27060

Question Number 27060 by math solver last updated on 01/Jan/18

Commented by math solver last updated on 02/Jan/18

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Answered by prakash jain last updated on 02/Jan/18

\frac{k - z_{1} {\bar{z}}_{2}}{z_{1} - {k z}_{2}} \times \frac{k - {\bar{z}}_{1} z_{2}}{{\bar{z}}_{1} - k {\bar{z}}_{2}} = 1

k^{2} + ∣ z_{1} ∣^{2} ∣ z_{2} ∣^{2} - k (z_{1} {\bar{z}}_{2} + z_{2} {\bar{z}}_{1})

=∣ z_{1} ∣^{2} + k^{2} ∣ z_{2} ∣^{2} - k (z_{1} {\bar{z}}_{2} + z_{2} {\bar{z}}_{1})

k^{2} + ∣ z_{1} ∣^{2} ∣ z_{2} ∣^{2} =∣ z_{1} ∣^{2} + k^{2} ∣ z_{2} ∣^{2}

∣ z_{2} ∣^{2} = 1