Find the number of values for m for which (m+i)^4 is an integer? i

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$i=sqrt(-1)$

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Answer 1 · 2017-11-16T17:00:35

$AA m in {0,+-1}, (m+i)^4 in ZZ.$

Using the Binomial Theorem, we have, $(m+i)^4,$

$=""_4C_0m^4i^0+""_4C_1m^3i+""_4C_2m^2i^2+""_4C_3m^1i^3+""_4C_4m^0i^4,$

$=m^4+4m^3i-6m^2-4mi+1,$

$rArr (m+i)^4=(m^4-6m^2+1) +(4m^3-4m)i.$

In order that $(m+i)^4$ be integer (real), equating the Real and

Imaginary parts of both sides, we must have,

$(m^4-6m^2+1) in RR, and 4m^3-4m=4m(m^2-1)=0.$

Now, $4m^3-4m=4m(m^2-1)=0 rArr m=0, m=+-1.$

$"Also, "AA m in {0,+-1}, (m^4-6m^2+1) in RR" holds good."$

$:. m in {0,-1,+1}.$

Hence, there are $3$ values of $m$ such that, $(m+i)^4 in ZZ.$