If a - ib - 1 - i-1 - i - then prove that a2 - b2 - 1

According to question, a+ib=1+i1 i On taking modulus on both side we get, |a+ib|=|1+i||1 i| = gt; #160; √(a^2+b^2)=11 = gt; #160; a^2+ ...

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Complex Numbers

If a + ib =...

Question

If $(a + i b) = \frac{1 + i}{1 - i}$ , then prove that $(a^{2} + b^{2}) = 1$

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Solution

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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}

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

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

(a + i b) = \frac{(x + i)^{2}}{(2 x^{2} + 1)}

(a^{2} + b^{2}) = \frac{(x^{2} + 1)^{2}}{(2 x^{2} + 1)^{2}}

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

c

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

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

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