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Properties of Modulus

Let 7+2in=a...

Question

Let $(7 + 2 i)^{n} = a + i b$ then show that $53^{n} = a^{2} + b^{2}$ .

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Solution

Given,
$(7 + 2 i)^{n} = a + i b$ .
Taking modulus both sides we get,
$| (7 + 2 i)^{n} | = | a + i b |$
$| (7 + 2 i) |^{n} = | a + i b |$ [ Since $| z^{n} | = | z |^{n}$ ]
or, $(\sqrt{53})^{n} = \sqrt{a^{2} + b^{2}}$
Now squaring both sides we get,
$53^{n} = a^{2} + b^{2}$ .

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