If frac{c + i}{c - i} = a + ib,where a, b, c | vedclass

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(A) Given $\frac{c + i}{c - i} = a + ib$  $(i)$Taking the conjugate on both sides,we get $\frac{c - i}{c + i} = a - ib$  $(ii)$Multiplying equation $(

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(A) Given $\frac{c + i}{c - i} = a + ib$  $(i)$Taking the conjugate on both sides,we get $\frac{c - i}{c + i} = a - ib$  $(ii)$Multiplying equation $(i)$ and equation $(ii)$,we get:$\left(\frac{c + i}{c - i}ight) 	imes \left(\frac{c - i}{c + i}ight) = (a + ib)(a - ib)$$1 = a^2 - (ib)^2$$1 = a^2 - i^2b^2$Since $i^2 = -1$,we have $1 = a^2 + b^2$Therefore,$a^2 + b^2 = 1$.

If $\frac{c + i}{c - i} = a + ib$,where $a, b, c$ are real,then $a^2 + b^2 = $

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