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Homework Help: Converting to Polar and Cartesian form

  1. Oct 16, 2014 #1
    Mod note: This post with template not used and no effort shown received a warning.

    Okay I am totally confused in this.
    This is not a homework question but rather one I saw online and was wondering for example how to solve it

    The question was -3-i/-8+6i to be expressed into Cartesian form.

    Another similar one was -2-2*sqrt(3i) into polar form.

    How can I do this?

    Can someone actually explain I don't get it fully. I know you usually use Cartesian form but yeah...
    Last edited by a moderator: Oct 17, 2014
  2. jcsd
  3. Oct 16, 2014 #2

    Simon Bridge

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    Do you know what polar and cartesian forms are?
    Do you know how to find the real and imaginary parts of a complex number?
    Do you know how to find the modulus and argument of a complex number?
    If the answer to any of the above is "no", then please look them up before returning.
    If the answer to all of the above is "yes", then what's the problem?
    Please show your best attempt.

    1. search for: "rationalize the denominator"
    2. search for: "square root of pi"
  4. Oct 17, 2014 #3
    Are these complex numbers or vector components. When you have an i in the denominator, you may treat it as i^-1 which equals i which will help eliminating complex numbers. If it is a vector component, recall x=rcostheta y=rsintheta.
  5. Oct 17, 2014 #4


    Staff: Mentor

    It's fairly obvious that they are complex numbers.
    This doesn't help if the denominator has a sum of terms in it.
  6. Oct 17, 2014 #5

    Simon Bridge

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    ... misplaced a minus sign I think, typo?
    Just so people who google here don't get the wrong idea:$$\frac{1}{\sqrt{-1}}=\frac{1}{i}=-i$$ ... it's actually easier to use in the fraction form.

    The trick: Since I can multiply by 1 without changing the number, and any number divided by itself is 1, I can do: $$\frac{1}{i} = \frac{i}{i}\frac{1}{i} = \frac{i}{-1} = -i$$ ... see what I did there?

    The general form of this trick, when it leaves a real number in the denominator, is called "rationalizing the denominator" (see note post #2).
    Hopefully OP will get back to us and supply the needed information.
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