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Complex numbers in trigonometry form

  1. Oct 26, 2014 #1
    1. The problem statement, all variables and given/known data
    Write down number [tex]1+i[/tex] and [tex]1+i\sqrt{3}[/tex] in trigonometry form.



    2. Relevant equations
    For complex number [tex]z=x+iy[/tex]
    [tex]\rho=|z|=\sqrt{x^2+y^2}[/tex]
    [tex]\varphi=arctg\frac{y}{x}[/tex]
    And



    3. The attempt at a solution
    Ok. For [tex]z=1+i[/tex]
    [tex]\rho=\sqrt{1+1}=\sqrt{2} [/tex]
    [tex]\varphi=arctg\frac{y}{x}=arctg1=\frac{\pi}{4}[/tex]
    So
    [tex]1+i=\sqrt{2}(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4})
    or for
    1+i\sqrt{3}
    [tex]\rho=\sqrt{1+3}=2[/tex]
    [tex]\varphi=arctg(\sqrt{3})=\frac{\pi}{3}[/tex]
    What is easiest way to find arctg?
     
  2. jcsd
  3. Oct 26, 2014 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You could always use a calculator! But if you mean of "easy angles", such as [itex]\pi/4[/itex], [itex]\pi/3[/itex], and [itex]\pi/6[/itex], it helps to have some experience with such values. For 1+ i, for example, you can mark the point in the complex plane and imagine a right triangle with one leg, along the real-axis, of length 1 and the other, perpendicular to the x-axis, of length 1. That is obviously an isosceles right triangle which tells you that the two acute angle are equal. Since they must add to [itex]\frac{\pi}{2}[/itex], they must each be [itex]\frac{\pi}{4}[/itex].

    For [itex]1+ i\sqrt{3}[/itex], we have a right triangle with one leg, along the real axis, of length 1 and the other, perpendicular to the x-axis, of lenth [itex]\sqrt{3}[/itex]. You then get, as you did, that the hypotenuse of that right triangle is 2. Okay, imagine "flipping" that right triangle about the vertical leg. The leg of the new right triangle in the x-axis goes from 1 to 2 so the triangle made from both right triangle has base from 0 to 2, of length 2, the same as its other two sides. That is, the triangle is an equilateral triangle so all three of its angles are the same, each is [itex]\frac{\pi}{3}[/itex].
     
  4. Oct 26, 2014 #3
    I would always recommend drawing an Argand diagram when attempting to determine the length and angle of complex numbers - this is where your equations originate anyway. After calculation it also serves as a decent way to check your numeric answers.

    The length (modulus), ## \rho ##, may be found by using pythagoras as a right-angle triangle and the angle (argument), ## \phi ##, may be found by simple trigonometry.
     
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