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I'm currently reading some notes on Complex Numbers and came across this 'proof' regarding the equality of complex numbers.

Claim: Two complex numbers $a+bi$ and $c+di$ are equal iff $a=b$ and $c=d$, where $a,b,c,d$ are real numbers.

Now this is an obviously true statement, but I had a problem with the proof provided.

Proof provided

Suppose $a+bi=c+di$

$a-c=i(d-b)$

$(a-c)^2=-(d-b)^2$

$(a-c)^2+(d-b)^2=0$

This is only possible iff $a=c$ and $b=d$ and so both the forward and converse are true.


I can see that this indeed proves the forward statement, but I had issues with the converse. I'd be okay with this if every single step works both ways, but the third line involves squaring both sides, which unless I'm mistaken does not work both ways.

Is there indeed an issue with this proof?

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    $\begingroup$ All of the above works ONLY if $a,b,c,d$ are assumed to be real. I would have stated that explicitly. $\endgroup$ – Michael Hardy Aug 20 '14 at 16:03
  • $\begingroup$ I've added that a,b,c,d are all real, which I forgot to add before. Now, in its current form, is the provided proof correct? $\endgroup$ – Trogdor Aug 20 '14 at 16:06
  • $\begingroup$ Note that the converse assumes that $a = c$ and $b = d$ so by removing the squares you have $a-c = \pm i(d-b)$ so you'd be proving that either $a+bi = c+di$ or that $a+di = c+bi$ but since $b = d$ you have that $a+bi = c+di$ as desired. $\endgroup$ – Darth Geek Aug 20 '14 at 16:11
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The given proof is correct. The converse is simpler. If $a=c$ and $b=d$, the $a+ib=c+id$ simply because $a=c$ means $a$ and $c$ are the same thing, so any expression involving $a$ remains the same if $c$ is put in place of $a$, and similarly for $b$ and $d$.

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