Simplifying a quotient of complex numbers

Given the equation I am supposed to simplify :

$$\frac{(7 - 4i)}{(5 + 3i)}$$

I conclude that I should first multiply both the numerator and denominator by $(5 - 3i)$ (note : or by $7 + 4i$ but either will do), which leads me to :

$$\frac{(35 - 41i + 12i^2)}{(25 - 9i^2)}$$

However, none of the answers in the multiple choice answer sheet agree with my solution. I stare at the answer sheet for half an hour wondering how is it that none of the answers given contains a term containing $i^2$. Just looking at the problem I know that I should expect the last term to be the product of $-4i$ and $3i$ which should yield $-12i^2$. So I start pulling my hair out and eventually I find wolfram's algebra simplifier and I run the expression through them to get a step by step walkthrough and find that when they multiply the last terms for both numerator and denominator they ignore the i and invert the sign! For example, in the numerator their last term ends up being $-4i * -3i = -12$ (not $12i^2$ according to the rules I learned) and the denominator's last term ends up being $3i * -3i = 9$ (not $-9i^2$). Please, for all that's holy and sacred in the world of algebra, someone explain to me how this magic occurs because I'm about to throw myself out of the proverbial window!

Here's the answer which I don't understand for the life of me :

$$\frac{(23 - 41i)}{34}$$

p.s. If you could recommend me something, I am looking for an Algebra problem book which has challenging equations to simplify with steps and tips/tricks... I really want to nail this.

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Note that you generally want to remove $i$ from the denominator, so you multiply by the conjugate of the denominator using $(a+bi)(a-bi)=a^2-b^2i^2=a^2+b^2$ because $i^2=-1$. Using $7+4i$ would not do the same. – Mark Bennet Jun 1 '13 at 17:13
Your second fraction was typed with $25$ instead of $35$ - which I corrected. – Mark Bennet Jun 1 '13 at 17:20
you realize $i^2=-1$ right? – aopsfan Jun 1 '13 at 17:20
@aopsfan: I think if you read the answers and comments below, you'll find that the OP is certainly well aware of it by now. – Cameron Buie Jun 1 '13 at 17:23
I've had the following conversation with several students. They wonder what to do with $i^2$. I ask if they knew that $i$ is a square root of $-1$. The students says yes, of course. I say: That should tell you what $i^2$ is. I get a blank look. Then I ask if they know what $\sqrt{83}\cdot\sqrt{83}$ is, and I'm told they'd need a calculator for that. Then I say "Do you know what square roots are?" and the answer is "Yes." I explain that $\sqrt{83}\cdot\sqrt{83}$ is $83$, and I'm asked how I knew that without a calculator. This has happened at least three times. – Michael Hardy Jun 1 '13 at 17:50

Hint: $$i^2:=-1$$${}{}{}{}{}{}$

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omg... I'm an idiot. I was looking at the variable i as if it were a simple variable and not the "imaginary" number i. Thank you for such a quick answer, and please... someone shoot me now... – Anthony Jun 1 '13 at 17:13
@Anthony you should accept either Mark or Cameron's answer if they helped you – DanZimm Jun 1 '13 at 17:16
@Anthony: In the interests of full disclosure, I did the exact same thing when I first encountered imaginary numbers. No worries! (And no gunshots, either.) – Cameron Buie Jun 1 '13 at 17:17
@Anthony you're not an idiot! ;-) – amWhy Jun 1 '13 at 17:30
Don't beat yourself up. I was really impressed by the exposition of your question! You understand how math works very well. – user29743 Jun 1 '13 at 17:52

As others have noted, you forgot that $\;\color{blue}{ \bf i^2 = -1},\;$ and so your expression simplifies.

Note however, that your expression should have been: $$\frac{(\color{blue}{\bf 35} - 41i + 12i^2)}{(25 - 9i^2)}$$

(Perhaps that was a typo when formatting?)

Now, simplify, substituting $\;i^2 = -1$:

Your choice to multiply numerator and denominator by the conjugate of $5 + 3i$, which you know is $5 - 3i$ was the correct choice, to clear the imaginary number from the denominator. Had you multiplied numerator and denominator by the conjugate of $7- 4i$, you would clear the imaginary number from the numerator, but it would remain in the denominator.

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I just corrected that as a computational slip in the question - but I gave you a vote for noticing. – Mark Bennet Jun 1 '13 at 17:21
Thanks, @Mark: It didn't occur to me, until re-editing, that the error may have been simply in typing/formatting the intended computation. – amWhy Jun 1 '13 at 17:28
@amWhy: and I'll give you another! :-) +1 – Amzoti Jun 2 '13 at 2:55
Thanks, dear friend! – amWhy Jun 2 '13 at 3:00

This all works because $i^2$ is defined to be equal to $-1$

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