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If a+i b=(c+i)/(c-i) , where c is real, ...

If `a+i b=(c+i)/(c-i)` , where `c` is real, prove that:`a^2+b^2=1 and b/a=(2c)/(c^2-1)dot`

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To solve the problem, we start with the equation given: \[ a + ib = \frac{c + i}{c - i} \] where \(c\) is a real number. We will prove that \(a^2 + b^2 = 1\) and \(\frac{b}{a} = \frac{2c}{c^2 - 1}\). ...
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