Question

If (a+bi) (c+di) (e+if) (g+hi) =A+Bi
Show that:
(a^2+b^2) (c^2+d^2) (e^2+f^2) (g^2+h^2) =A^2+B^2

Answer 1

I hope it will help you .......

Answer 2

Explanation:

Given: $(a+ib) (c+di) (e+if) (g+hi)= A+Bi$

Show that: $(a^{2} +b^{2} ) (c^{2} +d^{2} ) (e^{2} +f^{2} ) (g^{2} +h^{2} )=A^{2} +B^{2}$

Solution: $(a+ib) (c+di) (e+if) (g+hi)= A+Bi$   -----(1)

Taking conjugate both sides

$\overline{(a+ib)}\overline{(c+di)}\overline{(e+fi)} \overline{(g+hi)}=\overline{A+Bi}$

$(a-bi) (c-di) (e-if) (g-hi) =A-Bi$ -----(2)

Multiply both equation (1) and (2)

$(a+bi)(a-bi)(c+di)(c-di)(e+if)(e-fi)(g+hi)(g-hi)=(A+Bi)(A-Bi)\\\\(a^{2} -(bi)^{2} ) (c^{2} -(di)^{2} ) (e^{2} -(fi)^{2} ) (g^{2} -(hi)^{2} )=A^{2} -(Bi)^{2} \\\\(a^{2} +b^{2} ) (c^{2} +d^{2} ) (e^{2} +f^{2} ) (g^{2} +h^{2} ) =(A^{2} +B^{2} )$

Hence, showed.

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