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Math Help - polar and cartesian coordinates

  1. #1
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    polar and cartesian coordinates

    express values of the following expressions in both polar and cartesian coordinates and plot them
    a) square root of i
    b) ((1+i)/square root of 2) to the 25 power

    thanks, the i are confusing me
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  2. #2
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    Quote Originally Posted by dixie View Post
    express values of the following expressions in both polar and cartesian coordinates and plot them
    a) square root of i
    b) ((1+i)/square root of 2) to the 25 power

    thanks, the i are confusing me
    1)  a+bi = \sqrt{i}

     (a+bi)^2 = i

     a^2+2abi -b^2 = i

     a^2-b^2 +2abi = i

    Compare imaginary parts  2abi = i

     2ab = 1

     ab = \frac{1}{2}

     a = \frac{1}{2b}

    Now compare real parts

    ( \frac{1}{2b})^2-b^2 = 0

     \frac{1}{4b^2}  = b^2

     \frac{1}{4}  = b^4

     \pm (\frac{1}{4})^{\frac{1}{4}}  = b

     \frac{\sqrt{2}}{2}  = b

     a = \frac{1}{2}.\frac{2}{\sqrt{2}}

     a = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}

     \therefore \sqrt{i} = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}

    b)  a+bi = (\frac{1+i}{\sqrt{2}})^{25}

     (a+bi)^{\frac{1}{25}} = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i)

     (a+bi)^{\frac{1}{25}} = (\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i)

    And the RHS, as we seen in question a), is  \sqrt{i}

     (a+bi)^{\frac{1}{25}} =\sqrt{i}

     a+bi =\sqrt{i^{25}}

    A quick check on the calculate says that  i^{25} = i .

     a+bi =\sqrt{i} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i

    Both were identical. I'll leave you to put them in polar form
    Last edited by Mush; January 7th 2009 at 08:55 PM.
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