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Can someone help me with this simple problem

  1. Jun 18, 2003 #1
    Express (i-3) to the 4th power in simplest a+bi form.

    I know the answer , but cannot get my answer to be correct. Can someone help me solve this?

  2. jcsd
  3. Jun 18, 2003 #2
    Howdy Wasper,

    Ok, I'd break it down like this;

    (i-3)(i-3) = i^2-6i+9

    Now, because the original is to the fourth power you would have two groups like this;


    Multiply all the above out properly and after combining like terms you have;

    = i^4-12i^3+54i^2-108i+81

    Now, using what you know about 'i' raised to a power (how it repeats, and so forth) you will be able to transform some i's into numbers. The numbers will then combine and reduce to;


    Well, at least I think that's about right, haha, because I think I goofed on one of the signs but I'm too lazy to go back over it again and check my work. I think you can take it from here. It's actually quite a fun thing to do when you start getting the hang of it and have your mind free to concentrate.

    Ooops, I just noticed this was homework help :frown:
    Anyway, since I've gone this far already and because I don't know your level of understanding I'll rehash what I did above a little;

    (i-3)^4 = (i-3)(i-3)(i-3)(i-3)

    I broke it down into two groups of;
    [(i-3)(i-3)]x[(i-3)(i-3)] = [i^2-6i+9]x[i^2-6i+9]

    Do you follow?
    You could have just kept grunting them out by only multiplying the quantity (i-3) over and over, but the way I did it seemed more of a shortcut (but it is probably easier to confuse yourself doing it that way too). Lastly, the imaginary part might seem confusing at first but is actually rather easy if you beat it into your brain that because i = squareroot of negative one you get the following repeating patern;

    i^1 = sqrt -1
    i^2 = (sqrt -1)(sqrt -1) = -1
    i^3 = (-1)(sqrt -1) = (-1)(i) = -i
    i^4 = (-1)(-1) = 1

    Looking at the results from above you can see that i^5 is just going to be (sqrt -1)(1) = sqrt -1 = i^1, so i^5 ends up being i^1 (pattern repeats all over again). Do you suppose the odds are high that i^6 will end up being equivalent to i^2 ?? Yes, because it would be the equivalent of (i^2)(i^4) = (-1)(1) = -1 = i^2.

    On i^4, an easy way to determine the answer is is by recalling what i^2 was equal to). You could grunt it out, however, by using the value of i^3 multiplied by sqrt -1 and getting in lots of good practice manipulating things;

    = (-i)(sqrt -1)
    = (-i)(i)
    = -i^2
    = (-i)(-i)
    = (-sqrt -1)(-sqrt -1)
    = -[(sqrt -1)(sqrt -1)]
    = -[-1]
    = 1

    There, I hope that makes up for my having blabbed to much.
    Last edited by a moderator: Jun 19, 2003
  4. Jun 19, 2003 #3


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    Staff Emeritus
    Science Advisor

    Just to point out that you don't really NEED to multiply
    (i^2- 6i+ 9)(i^2- 6i+ 9) in that form.

    i^2- 6i+ 9 is, of course, -1- 6i+ 9= 8- 6i so it is simpler
    to multiply (8- 6i)(8- 6i)= 64- 2(8)(6)i+ 36i^2= 64- 96i- 36
    = 28- 96i just as BoulderHead said.
  5. Jun 19, 2003 #4
    my mistake was trying to do the (i-3) square it to the 4th power all at the same time. I think I got it now.. thanks for the help guys.
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