Question

3. If (p+iq) ^2 = x+iy then prove that (p^2+q^2)^2= x + y?​

Answer 1

$Given \: \blue{(p+iq)^{2} = x + iy }$

$\implies p^{2} + (iq)^{2} + 2 \times p \times iq = x + iy$

$\implies p^{2} - q^{2} + 2 i pq = x + iy$

$\boxed{ \pink{ \because i^{2} = -1 }}$

$Compare \: both \:sides, we \:get$

$p^{2} - q^{2} = x \: --(1)$

$2pq = y \: --(2)$

$LHS = \red{ (p^{2} + q^{2})^{2} }$

$= (p^{2} - q^{2})^{2} + 4p^{2}q^{2}$

$= (p^{2} - q^{2})^{2} + (2pq)^{2}$

$\green {= x^{2} + y^{2}} \: \blue{ [ From \: (1) \:and \:(2) ]}$

$\neq RHS$

Therefore.,

$\red{ (p^{2} + q^{2})^{2} }\green { = x^{2} + y^{2}}$

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