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If $m$ and $n$ are real numbers, $i^2 =-1$, and $(m-n)-4i=7+ni$, what is the value of $m$?

2 Answers 2

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Two complex numbers are equal if and only if their real and imaginary parts are equal. From this you can deduce that $n = -4$ and $m-n = 7$. Now plug $n=-4$ into the second equation and solve for $m$.

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    @Neal: To make more explicit: The left hand side is $(m-n) + (-4)i$ and the right hand side is $7+ n i$. Since both $m$ and $n$ are real by assumption, so is $m-n$, hence both $(m-n) + (-4)i$ and $7 + ni$ are of the form $a + bi$ with $a$ and $b$ real. If you're not familiar with complex numbers, you might want to have$a$look at the wikipedia page: http://en.wikipedia.org/wiki/Complex_number2011-03-27
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HINT $\: $ Suppose more generally that $\rm\:j\:$ is any irrational number and that $\rm\ (n+4)\ j\ =\ m-n-7\:.\:$ It follows that $\rm\ n+4 = 0\ $ for else we would have that $\rm\ j = (m-n-7)/(n+4)\ $ would be rational. Finally you can solve for $\rm\:m\:$ by substituting $\rm\:n= -4\:$ above. Note that, in your special case, $\rm\ j = i\ $ is irrational since it has negative square $\rm\ i^2 = -1\:,\:$ but every rational has nonnegative square. Hence the solution is the same for any irrational $\rm\:j\:,\:$ e.g. $\rm\ \sqrt{2},\ i,\ \pi\ $ and even for $\rm\: j = x\:$ an indeterminate (here $\rm\:j\:$ is irrational means that a $\rm\:j\:$ is a nonrational element in some ring containing the rationals).