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Show 5i/(2+i) = 1+ 2i

  1. Sep 17, 2011 #1
    1. The problem statement, all variables and given/known data

    Show this by writing the individual factors on the left in exponential form, performing the needed operations, and finally changing back to rectangular coordinates.

    2. Relevant equations

    i is an imaginary number.

    3. The attempt at a solution

    Looking at the numerator, z_1 = 5i where r = |z_1|= 5, theta = pi/2.
    Looking at the denominator, z_2 = 2+i where r = |z_2| = sqrt(5), theta = arctan(1/2).

    So, in exponential form, 5i/(2+i) becomes 5*e^(i*pi/2) / sqrt(5)*e^(i*arctan(1/2)) =>
    sqrt(5)*e^(i*pi/2) / e^(i*arctan(1/2)) = sqrt(5)*e^(i*((pi/2) - arctan(1/2))) but I don't see how this can be turned back into 1+2i since 1+2i in exponential form
    is sqrt(5)*e^(i*arctan(2)).

    Am I missing an algebra step or did I do something wrong?

    Thank you.
     
  2. jcsd
  3. Sep 17, 2011 #2

    dynamicsolo

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    You have used Euler's Identity and DeMoivre's Theorem for division correctly. The issue now is to use the right trig identity.

    So [itex] \frac {5 cis \frac{\pi}{2}}{\sqrt{5} cis \theta} = \sqrt{5} cis ((\frac{\pi}{2}) - \theta) [/itex] . You wish to find [itex]arctan( \frac{\pi}{2} - \theta )[/itex]. Make a picture of a right triangle with one angle being [itex]\theta[/itex] , having [itex]tan \theta = 1/2 [/itex]; then the other angle is the complement [itex]\frac{\pi}{2} - \theta[/itex]. Looking at that angle, what would its tangent be? That will give you the rectangular components you need.
     
  4. Sep 17, 2011 #3

    ehild

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    Gold Member

    Use that

    tan(alpha) = 1/tan(pi/2-alpha), that is, arctan(1/2)+arctan(2)=pi/2.


    ehild

    Edit: Dynamicsolo beat me...
     
  5. Sep 17, 2011 #4

    Ray Vickson

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    When simplifying complex fractions it is always a good idea to multiply and divide by a factor that makes the denominator real. Your method is WAY too complicated.

    RGV
     
  6. Sep 17, 2011 #5

    dynamicsolo

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    I don't think anyone would usually do the calculation by other than the "conjugate factor method", but to judge from sarahs52's post, this appears to be a "demonstration problem", where you're asked to show that the solution of the problem by an alternate method does indeed yield the same answer as the "usual" method would...
     
  7. Sep 22, 2011 #6
    Yes, dynamicsolo is correct. As I stated at the beginning of my post, the question asks to write "...the individual factors on the left in exponential form, performing the needed operations, and finally changing back to rectangular coordinates."

    Thanks everyone!
     
  8. Sep 26, 2011 #7
    I do not have the trigonometric breakdown, so this may not be very satisfying, but ((pi/2) - arctan(1/2)) is indeed equal to arctan(2). Once you show that equivalence, you can rely on transitivity to take you the rest of the way.
     
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