If a - i b - p - i 2 q - i then prove that a 2 - b 2 - p 2 - 1 4 q 2 - 1

Given that, a+ib=p+i2q i Taking Modulus on both sidesWe get |a+ib|=|p+i||2q i| √(a^2+b^2)=√(p^2+1)√(4q^2+1) Squaring on both sides, we get a^2 ...

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Modulus of a Complex Number

If a + i b ...

Question

If $a + i b = \frac{p + i}{2 q - i}$ then prove that $a^{2} + b^{2} = \frac{p^{2} + 1}{4 q^{2} + 1}$

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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{1 + i}{1 - i}

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

\frac{a + i b}{c + i d} = p + i q

\frac{a - i b}{c - i d} = p - i q

p^{2} + q^{2} = \frac{a^{2} + b^{2}}{c^{2} + d^{2}}

(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{{(x + i)}^{2}}{{2 x}^{2} + 1}

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

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