If a-ib-c-ic-i- where c is a real number- then

The correct option is A ba=2cc^2 1 Simplifying the above expression gives us #160; a+ib=(c+i)^2c^2+1 =c^2 1+i2cc^2+1 Hence , a=c^2 1c^2+1 A ...

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Algebra of Complex Numbers

If a+ib=c+i...

Question

If $a + i b = \frac{c + i}{c - i}$ , where $c$ is a real number, then

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

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

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\frac{b}{a} = \frac{4 c}{c^{2} - 1}

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none of these

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

c

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

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

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

a, b, c

a^{2} = b^{2} + c^{2} - 2 b c \sqrt{1 - a^{2}}, b^{2} = c^{2} + a^{2} - 2 c a \sqrt{1 - b^{2}}, c^{2} = a^{2} + b^{2} - 2 a b \sqrt{1 - c^{2}}

a = c \sqrt{1 - b^{2}} + b \sqrt{1 - c^{2}}

For every real number c $\geq 0$ , find all the complex number z which satisfy the equation

$| z |^{2}$ - 2iz + 2c(1 + i) = 0

a^{2} + b^{2} + c^{2} = 1, b + i c = (1 + a) z

\frac{a + i b}{1 + c} = \frac{1 + i z}{1 - i z}

a, b, c

z

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