If a - bi - c - i - ci - a - b - c - R- show that a 2- b 2-1 and b - a -2 c -c2-1

A+bi=c+i/c i=c+i/c i.c+i/c+i=c^2+2ci+i^2/c^2 i^2=c^2+2ci 1/c^2+1 =c^2 1/c^2+1+2c/c^2+1i Equating real and imaginary parts, we have a=c^2 1/c^2+1, b =2c/c^2+1 ∴ ...

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Byju's Answer

Standard XII

Mathematics

Equality of 2 Complex Numbers

If a + bi = c...

Question

If $a + b i = \frac{c + i}{c - i}, a, b, c \in R$ , show that $a^{2} + b^{2} = 1$ and $\frac{b}{a} = \frac{2 c}{c^{2} - 1}$

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Solution

$a + b i = \frac{c + i}{c - i} = \frac{c + i}{c - i} . \frac{c + i}{c + i} = \frac{c^{2} + 2 c i + i^{2}}{c^{2} - i^{2}} = \frac{c^{2} + 2 c i - 1}{c^{2} + 1} = \frac{c^{2} - 1}{c^{2} + 1} + \frac{2 c}{c^{2} + 1} i$
Equating real and imaginary parts, we have
$a = \frac{c^{2} - 1}{c^{2} + 1}, b = \frac{2 c}{c^{2} + 1} ∴ a^{2} + b^{2} = {(\frac{c^{2} - 1}{c^{2} + 1})}^{2} + {(\frac{2 c}{c^{2} + 1})}^{2} = \frac{(c^{2} - 1)^{2} + 4 c^{2}}{(c^{2} + 1)^{2}} = \frac{(c^{2} + 1)^{2}}{(c^{2} + 1)^{2}} = 1$
$\frac{b}{a} = \frac{2 c}{c^{2} - 1} = \frac{2 c}{c^{2} - 1}$

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a + i b = \frac{c + i}{c - i}

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If a+bi=c+ic−i,a,b,c∈R, show that a2+b2=1 and ba=2cc2−1

a+bi=c+ic−i=c+ic−i.c+ic+i=c2+2ci+i2c2−i2=c2+2ci−1c2+1=c2−1c2+1+2cc2+1i Equating real and imaginary parts, we have a=c2−1c2+1,b=2cc2+1∴a2+b2=(c2−1c2+1)2+(2cc2+1)2=(c2−1)2+4c2(c2+1)2=(c2+1)2(c2+1)2=1 ba=2cc2−1=2cc2−1

If $a + b i = \frac{c + i}{c - i}, a, b, c \in R$ , show that $a^{2} + b^{2} = 1$ and $\frac{b}{a} = \frac{2 c}{c^{2} - 1}$