{"id":845,"date":"2015-06-17T09:00:34","date_gmt":"2015-06-17T09:00:34","guid":{"rendered":"https:\/\/magoosh.com\/act\/?p=845"},"modified":"2017-04-28T12:35:36","modified_gmt":"2017-04-28T19:35:36","slug":"imaginary-numbers-on-the-act","status":"publish","type":"post","link":"https:\/\/magoosh.com\/act\/imaginary-numbers-on-the-act\/","title":{"rendered":"Imaginary Numbers on the ACT"},"content":{"rendered":"

Sadly, ACT imaginary numbers aren\u2019t the ones you may have come up with as a child, like eleventy-twelve.<\/em> (Though if any of you become mathematicians one day, please<\/em> make that happen. It would be fun!)<\/p>\n

No, real-life imaginary numbers (and isn\u2019t that<\/em> a weird turn of phrase) were discovered\/invented as a way to take the square root of a negative number. With real<\/em> numbers, we can\u2019t do that, but by using our imaginary number, we totally can!<\/p>\n

Basically, if you take the square root of -1, you wind up with the imaginary number i<\/em><\/p>\n

Translated into math, it looks like this:<\/p>\n

\"in_img1\"<\/p>\n

Simple so far, yeah? Good, because we\u2019re about to introduce a little complexity up in here.<\/p>\n

 <\/p>\n

Simplifying Imaginary Numbers on ACT Math<\/h2>\n

Okay, let\u2019s take a look at an example. Say you needed to find the square root of -16.<\/p>\n

\"in_img2\"<\/p>\n

Well, how do you get the number -16? One way is to multiply 16 and -1, right? So let\u2019s rewrite it like this:<\/p>\n

\"in_img3\"<\/p>\n

We already know that \u221a-1 is i<\/em>, and you should know that \u221a16 is 4. So now our example looks like this:<\/p>\n

\"in_img4\"<\/p>\n

And there\u2019s our answer!<\/p>\n

Simple, right? No? Okay then, let\u2019s try another example. What if we wanted to find \u221a-x? Using the same steps we followed above, here\u2019s our progression:<\/p>\n

\"in_img5\"<\/p>\n

A little more complicated now, but still totally doable. Your answer would be i<\/em><\/strong>\u221ax<\/strong>.<\/p>\n

 <\/p>\n

Powers of Imaginary Numbers<\/h2>\n

All right, stop looking at me like that. This is way<\/em> easier than it sounds at first.<\/p>\n

Let\u2019s say you needed to figure out what \"i^2\" <\/sup>is , or, to write it another way i x i<\/em>. Since we already know that i<\/em> is \u221a-1, asking for \"i^2\" <\/em>is just another way of saying \"{(sqrt{-1})}^2\" . When you square a square root, both of them cancel out, leaving you with the answer of -1<\/strong>.<\/p>\n

So, \"{i^2}=.<\/p>\n

How about \"i^3\" <\/sup>? That would be . <\/em>But instead of thinking of it as (\u221a-1)*(\u221a-1)*(\u221a-1), think of it as building on <\/em>the power we already know, namely . <\/em>So \"i^3\" would look like this:<\/p>\n

\"in_img6\"<\/p>\n

which is just another way of saying<\/p>\n

-1 x i<\/em><\/p>\n

which is just another way of saying<\/p>\n

-i<\/em><\/p>\n

So, \"{i^3}=.<\/p>\n

Easy peasy so far, right? Let\u2019s try \"{i^4}=. We can break that down like so:<\/p>\n

\"in_img8\"<\/p>\n

As we already know, \"{i^2}\" is -1. This is just a fancy way of writing<\/p>\n

\"in_img9\"<\/p>\n

which, of course, equals 1<\/strong>.<\/p>\n

\"{i^4}=<\/p>\n

Now let\u2019s live dangerously and figure out \"{i^5}\". And remember, we\u2019re building on what we already know, so we can rewrite it like this:<\/p>\n

\"{i^4}<\/p>\n

We already know that \"{i^4}=, so our example now looks like this:<\/p>\n

\"1<\/p>\n

Or, to simplify:<\/p>\n

i<\/em><\/p>\n

\"{i^5}<\/p>\n

A Nifty Little Trick<\/h2>\n

Let\u2019s write out everything we\u2019ve learned so far in a list, including \"{i^1}\"<\/p>\n

\"in_img10\"<\/p>\n

Take a close look at everything we\u2019ve found so far. The powers of i<\/em> work in a very clear cycle. You can remember it with a lovely little mnemonic device that I learned in school. Say the answers we uncovered aloud, ignoring the negative signs.<\/p>\n

\"in_img11\"<\/p>\n

All you have to do is remember \u201cI Won,\u201d and you can figure out any power of i<\/em>, any time, anywhere.<\/p>\n

But wait!<\/em> I hear you cry. What about the negative signs?<\/em> Well, just remember that the negative signs are on the inside<\/em> of the cycle, like so:<\/p>\n

\"in_img12\"<\/p>\n

Now that we\u2019ve reviewed all that, let\u2019s look at an ACT-style example.<\/p>\n

The imaginary number i<\/em> is defined such that \"{i^2). What does \"in_img13\"<\/p>\n

equal?<\/p>\n

    \n
  1. i<\/li>\n
  2. -i<\/li>\n
  3. 0<\/li>\n
  4. -1<\/li>\n
  5. 1<\/li>\n<\/ol>\n

    I\u2019ll give you a minute to try to puzzle this one out. If you need me, I\u2019ll be humming the song they play on \u201cJeopardy!\u201d<\/p>\n

    .<\/strong><\/p>\n

    .<\/strong><\/p>\n

    .<\/strong><\/p>\n

    Okay, how did you do? Did you remember \u201cI -Won, -I Won\u201d? Let\u2019s take a look.<\/p>\n

    The first four terms in our question work out like this:<\/p>\n

    i + <\/em>-1 + -i <\/em>+ 1 = 0<\/p>\n

    You can count on the sum being 0<\/strong> every four terms<\/strong>, since we\u2019re working with a cycle. So let\u2019s think about this logically. We\u2019re dealing with \"{i^21}\" at the end of our question. How many groups of four are there in 21?<\/p>\n

    \"in_img14\"<\/p>\n

    As far as we\u2019re concerned, that 0.25 left at the end is all we care about. That means we\u2019re one term into the next cycle. And if you remember your mnemonic device, we\u2019ve gone through \u201cI -Won, -I Won\u201d five times in this problem. As the first term is always i, <\/em>our answer is A. <\/strong><\/p>\n

    Here\u2019s one to try on your own:<\/p>\n

    The imaginary number i<\/em> is defined such that \"{i^2) . What does \"in_img15\"equal?<\/p>\n

      \n
    1. i<\/li>\n
    2. -i<\/li>\n
    3. 0<\/li>\n
    4. i-1<\/li>\n
    5. 1-i<\/li>\n<\/ol>\n

      Remember our cycles of four. How many of those are we dealing with? 54\/4 = 13.5<\/p>\n

      Disregard the first 13 cycles and just look at the 0.5 at the end. That means we\u2019re two terms into the 14th cycle, and everything before that cancels out. i^53 = i and i^54 = -1, so the remaining sum is i-1, so our answer is D<\/strong>. Remember \u201cI -Won, -I Won,\u201d and it’s easy!<\/p>\n

      For more on imaginary numbers, check out Kristin’s explanation below!
      \n 
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