If a-ib-p-i2q-i- prove that a2-b2-p2-14q2-1

a+ib = p+i2q i Taking modulus on both sides | a+ib | = | p+i2q i | a^2+b^2 = p^2+14q^2+1 ...

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Complex Numbers

If a+ib=p+i...

Question

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

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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{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}}

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

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

If a + ib =, prove that a² + b² =

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