Question

if a+ib= c+i / c-i , where c is real part prove that a square + b square =1 and b / a =2c / c square-1

Answer 1

this is the solution of ue problem

Answer 2

$\blue { a }+ i \pink {b} = \frac{c+i}{c-i} \: (given )$

$= \frac{(c+i)(c+i)}{(c-i)(c+i)} \\= \frac{(c+i)^{2}}{c^{2} - i^{2} } \\= \frac{c^{2} + i^{2} + 2 \times c \times i }{c^{2} + 1 }$

$= \frac{c^{2} - 1 + 2ic }{ c^{2} + 1 } \\= \blue {\frac{(c^{2}-1)}{c^{2}+1} }+ \pink {\frac{2c}{c^{2}+1}} i$

/* Compare bothsides ,we get */

$\blue {a = \frac{(c^{2}-1)}{c^{2}+1} } ,\pink {b = \frac{2c}{c^{2}+1}}$

$\red{ a^{2} + b^{2} } \\= \Big( \frac{c^{2}-1}{c^{2}+1}\Big)^{2} +\Big( \frac{2c}{c^{2}+1}\Big)^{2}$

$= \frac{ c^{4} - 2c^{2}+1 + 4c^{2}}{ (c^{2}+1)^{2}}$

$= \frac{ c^{4} +2c^{2}+1 }{ (c^{2}+1)^{2}}$

$= \frac{ (c^{2} + 1 )^{2}}{ (c^{2}+1)^{2}}\\= \green {1} \: --(1)$

$\red{ ii ) \frac{b}{a} } \\= \frac{\frac{2c}{c^{2}+1}} {\frac{(c^{2}-1}{c^{2}+1}}\\= \green {\frac{2c}{c^{2} - 1 } } \: --(2)$

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