If c - i - c - i - a - ib- where a - b - c are real- then a 2- b 2-A- -1B- -c2C- 1D- c2

The correct option is C 1 c+i/c i = a+ib .........(i) ∴ c i/c+i = a ib .........(ii) Multiplying (i) and (ii), we get c^2+1/c^2+1 = a^2 + b^2 ...

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Algebra of Complex Numbers

If c + i / c ...

Question

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

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Similar questions

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

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

c

a^{2} + b^{2} = 1

\frac{b}{a} = \frac{2 c}{c^{2} - 1}

a, b, c

d

a^{2} + b^{2} + c^{2} + d^{2} = 1

A = [\begin{matrix} a + i b & c + i d - c + i d & a - i b \end{matrix}],

i^{2} = - 1,

A^{- 1} =

Find:
$(a + b + c) \times (a + b - c) =$

\frac{b^{2} + c^{2}}{b + c} + \frac{c^{2} + a^{2}}{c + a} + \frac{a^{2} + b^{2}}{a + b} \geq a + b + c

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