If a-iba-ib-1-i1-i- then prove that a-b-0

We have, #10; #10; a iba+ib=1+i1 i #10; #10; ( a ib )( 1 i )=( a+ib #10;)( 1+i ) #10; #10; a ai ib+i^2b=a+ai+ib+i^2b #10; #10 ...

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Complex Numbers

If a-iba+ib...

Question

If $\frac{a - i b}{a + i b} = \frac{1 + i}{1 - i}$ , then prove that $a + b = 0$

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Solution

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(a + i b) = \frac{1 + i}{1 - i}

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

a + i b = \frac{p + i}{2 q - i}

a^{2} + b^{2} = \frac{p^{2} + 1}{4 q^{2} + 1}

a + i b = \frac{p + i}{2 q - i}

a^{2} + b^{2} = \frac{(p^{2} + 1)}{4 q^{2} + 1}

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

c

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

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

{(\frac{1 + i}{1 - i})}^{3} - {(\frac{1 - i}{1 + i})}^{3} = A + i B

A, B =

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