# Math question

The following appeared in The Seattle Times [reformatted for the message board]:

1. Write (1 - 3i) ÷ (-2 - 4i) in the form a + bi.

Simple algebra, I know. But it’s been a while. I haven’t worked on it, but at a glance I see that a 2 can be factored out of the denominator. I could multiply either the numerator or the denominator by its compliment, creating a perfect square. I’m otherwise occupied at the moment and don’t have the time to play with it right now, so I thought I’d post here to A) see if I’m on the right track; and B) to let others join in the fun.

1/2 + (1/2)i

There are two other questions. The first one is solvable at a glance. The second (which is the third question) I haven’t tried yet.

1. Solve for x: 8[sup]x[/sup] = 32

As I said, it’s solvable at a glance: 5/3

1. The graph of the equation y = 3[sup]x[/sup] is reflected over the y-axis. What is the equation of the image?

y = (1/3)[sup]x[/sup]

Everything you’ve said is correct.

(Except “The second (which is the third question) I haven’t tried yet”. Seems to me you’ve answered all the questions…)

Oh, wait, I see. You’ve read the answers and are posting them all, but don’t know how to obtain them. Heh, sorry, I misinterpreted you.

I think you mean “conjugate” rather than “compliment”, but other than that, yes. The standard MO for finding the real and imaginary part of (a + bi)/(c + di) is to multiply top and bottom by (c - di). The denominator becomes real, and you can multiply out the numerator to get the real and imaginary parts.

Question [1] isn’t simple algebra. i is an imaginary number squareroot(-1). But the solution is simple in that you

[spoiler]multiply the top and bottom by the conjugate of the bottom -2+4i
where i * i = -1
(1 - 3i) * (-2 +4i) = -2 +4i +6i +12 = 10 +10i

(-2 -4i) * (-2 +4i) = 4 -8i + 8i +16 = 20

1/2 + (1/2)i[/spoiler]

The third is probably the easiest - if you’re reflecting across the y axis, for any value of the function at x, the reflection will have the value at -x. Thus for r the reflection, r(x) = f(-x).

‘Conjugate’. I said it’s been a while. :smack:

That’s what I thought. I’m still elsewhere, but I’ll work through it after a while.

emacknight: Yes, I assumed people would know that i is the square root of negative one.

Incidentally, this appeared in an article written originally for The Washington Post about whether it would be a good idea to require Algebra II for all high school students:

When I teach people how to work that first problem, I tell them that we’re doing the same trick we do when we try to rationalize a denominator that contains two terms (which at worst contain square roots). Because the problem in both sorts of problems is that you have a square root in the denominator that you don’t want there; with a complex number, the square root you’re trying to get rid of is the square root of -1 (aka i). So, you just do the same thing.

Trying to write (1 - 3i) ÷ (-2 - 4i) in the form a + bi involves what I call “real-izing” the denominator, by which I mean changing the denominator into a real number. That’s no harder than trying to simply (or rationalize the denominator of) (1 - 3sqrt(2)) ÷ (-2 - 4sqrt(2)), where I’m using sqrt(2) to represent the square root of 2. In both problems, you multiply the numerator and denominator by the conjugate of the denominator. So, anybody who can successfullly rationalize denominators shouldn’t have much trouble with dividing a complex number by a complex number to get a number written in standard complex number form.

Now, for the 8^x = 32 problem, what we teach in algebra classes is that if a > 0 and a <>1, then a^b = a^c if and only if b = c. So, it’s just a matter of writing both sides of the equation as 2 raised to a power and then equating the powers. Thus: since 8 = 2^3 and 32 = 2^5, we have (2^3)^x = 2^5, or in other words 2^3x = 2^5, so 3x = 5, so x = 5/3.

The work for the other problem has already been described.