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I have a complex quotient $\frac{(3+i)^2}{(1+2i)^2}$

The solution provided in my textbook is $-2i$. I arrived at different solutions and I'd like to know where I went wrong.

Till now in my textbook chapter I have been working with the complex number i ($\sqrt{-1}$).

I understand that one cannot divide by a complex number in the denominator so I must multiply both the numerator and denominator by the complex conjugate.

However, for this exercise I'm confused since my expression is nested inside parenthesis and is squared. So, if for example my denominator was just $1+2i$ I know that the complex conjugate would be $1-2i$.

So I'm confused about what do do since the whole denominator is within parenthesis and squared.

Just using what I know I tried solving for the squared term in both numerator and denominator:

$(3+i)^2$ = $3^2+i^2$ = $9-1$ = $8$

For the denominator: $(1+2i)^2$ = $1^2+2^2i^2$ = $1+4 * -1$ = $1-4$ = $-3$

Then I would arrive at $\frac{8}{-3}$ which is not the solution.

How can I arrive at $-2i$?

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  • $\begingroup$ Square and simplify the denominator first, then find the conjugate of the result. I see that you attempted to do that, but you got the wrong result. Use $(a+b)^2=a^2+2ab+b^2$. $\endgroup$
    – John Wayland Bales
    Apr 19, 2019 at 15:38
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    $\begingroup$ $(a+b)^2$ is not simply $a^2+b^2$. This is such a common mistake that it has it's own name and wiki page. See The Freshman's Dream on wikipedia. Instead, $(a+b)^2 = a^2+2ab+b^2$. So, you should instead have had $(1+2i)^2 = 1^2 \color{red}{+2\cdot 1\cdot 2i} + (2i)^2$ and similarly for the numerator and you should be able to continue from there. $\endgroup$
    – JMoravitz
    Apr 19, 2019 at 15:40
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    $\begingroup$ As an aside, thank you for including what you have tried as it makes it much clearer where your misconceptions lie. +1 $\endgroup$
    – JMoravitz
    Apr 19, 2019 at 15:41
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    $\begingroup$ This is somewhat of an aside, but please, try to stay away from $i=\sqrt{-1}$ as much as you can. It's supposed to be $i^2=-1$. The difference between the two is bigger than one might at first suspect. Mainly, using the former lures new students into a false sense of security when it comes to square roots and complex numbers. But there are other, more subtle reasons too. $\endgroup$
    – Arthur
    Apr 19, 2019 at 15:46
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    $\begingroup$ And, possibly another aside, and a semantic one at that, but just to be sure that you understand... You wrote, "I understand one cannot divide by a complex number [...] " - Well, one can, and you did... However, as you clearly understand, one can give the result in a form with no complex number in the denominator (by fooling around with complex conjugates, say) - and this is [no doubt] what is desired. For instance $1/i = -i / 1= -i$, so $1$ divided by $i$ equals $-i$. $\endgroup$
    – peter a g
    Apr 19, 2019 at 15:56

3 Answers 3

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You accidentally fell into the trap of using The Freshman's Dream.

Exponentiation does not distribute over sums. In most every context $(a+b)^n$ is not equal to $a^n + b^n$ and in particular $(a+b)^2$ is not simply $a^2+b^2$ but is instead $a^2+2ab+b^2$. For larger exponents, see the binomial theorem.

In your attempt you tried to calculate the numerator as $(3+i)^2 = 3^2+i^2$ which is incorrect. Rather, you should remember to use the FOIL method correctly or learn the correct outcome. Instead, it should have been $(3+i)^2 = 3^2\color{red}{+2\cdot 3\cdot i}+i^2 = 9+6i-1=8+6i$.

You made a similar mistake for the denominator. Once those mistakes are corrected, you should be able to continue on your own using the method you allude to in your post of multiplying both top and bottom by the complex conjugate of the denominator and simplifying after.

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  • $\begingroup$ Thank you for your advise and letting me know about the freshman's dream!, I was able to arrive at the solution by correctly factoring out (if that's the phrase? reverse factoring, if you will) the denominator per FOIL $\endgroup$
    – Doug Fir
    Apr 19, 2019 at 15:57
  • $\begingroup$ The usual word to describe the process you are referring to is "expanding." $\endgroup$
    – JMoravitz
    Apr 19, 2019 at 15:58
  • $\begingroup$ Or "multiplying out" - but probably "expanding" is better... $\endgroup$
    – peter a g
    Apr 19, 2019 at 16:00
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Expanding we get $$\frac{9-1+6i}{1-4+4i}=\frac{8+6i}{-3+4i}$$ and now multiply numerator and denominator by $$-3-4i$$

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It might help to note that this expression is $$\left(\frac{3+i}{1+2i}\right)^2=\left(1-i\right)^2=-2i$$ assuming that you know how to divide $3+i$ by $1+2i$.

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