If a-i bc-i de-i fg-i h- A -i B- then show thata2-b2c2-d2e2-f2g2-h2- A 2- B 2

On squaring both sides, we obtain (a2 + b2) (c2 + d2) (e2 + f2) (g2 + h2) = A2 + B2 Hence, proved. ...

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Cube Root Using Prime Factorisation

If a+i bc+i d...

Question

If (a + ib) (c + id) (e + if) (g + ih) = A + iB, then show that

(a² + b²) (c² + d²) (e² + f²) (g² + h²) = A² + B².

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Solution

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If (a+ib)(c+id)(e+if)(g+ih)=A+iB, then ( $a^{2}$ + $b^{2}$ )( $c^{2}$ + $d^{2}$ )( $e^{2}$ + $f^{2}$ )( $g^{2}$ + $h^{2}$ )=

(a + i b) (c + i d) = A + i B

(a^{2} + b^{2}) (c^{2} + d^{2}) = A^{2} + B^{2}

A = [\begin{matrix} a + i b & c + i d - c + i d & a - i b \end{matrix}]

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

A^{- 1}

\frac{(a + i b)}{(c + i d)} = x + i y

\frac{a - i b}{c - i d} = x - i y

\frac{a^{2} + b^{2}}{c^{2} + d^{2}} = x^{2} + y^{2}

A = [\begin{matrix} a + i b & c + i d - c + i d & a - i b \end{matrix}]

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

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