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I want help to find tan of polar correctly

  1. Apr 19, 2010 #1
    Hi



    I face some problem when i try to find tan in like this questions



    for example : write in polar

    z1 = 1 - i

    the answer is -5π/4

    but my answer is

    tan Q = y/x = -1/1 = -1



    how I get -5π/4



    --------------



    ex : 2



    z = -root3 + i

    my answer is

    tanQ = -1/root3 = -0.143

    but the answer is = 5π/6



    How i get the answer 5π/6



    ----





    please help me >>
     
  2. jcsd
  3. Apr 19, 2010 #2
    A complex number on the form [tex]x+iy[/tex] and the point of your exercise is to find [itex]\theta[/itex]

    you know that

    [tex]tan(\theta) = \frac{y}{x}[/tex] Understand?
     
  4. Apr 19, 2010 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Do you even know what question you are trying to answer?
    "[itex]- 5\pi/4[/itex]" is NOT the answer to the question posed, to write 1- i in polar form but it is part of the answer. Yes, tan(Q)= -1 and a calculator will tell you that [itex]Q= tan^{-1}(-1)= -45[/itex] degrees or [itex]-\pi/4[/itex] radians. You should also be able to recognize that tangent is 1 when the "near side" and "opposite" side are the same- an isosceles right triangle with angles or 45 degrees= [ Of course, that is in the
    fourth quadrant where x is positive and y is negative so that is the correct answer. You can also write that, as a positive number, [itex]2\pi- \pi/4= 8\pi/4- \pi/4= 7\pi/4[/itex]. [itex]-5\pi/4[/itex] is NOT the correct argument. That is in the second quadrant and so is
    the argument for -1+ i.

    To find the polar form you will also need the modulus or absolute value. That is [math]\sqrt{1^2+ (-1)^2}= \sqrt{2}[/math].

    Once again, the value you are looking for is NOT "tan(Q)" but Q itself. To find that take the inverse tangent. A calculator will tell you that [itex]Q= tan^{-1}(-1/\sqrt{3})= -30 degrees= [itex]-\pi/6[/itex] radians. Again that is in the fourth quadrant (a calculator always gives arctan values between [itex]-\pi/2[/itex] and [itex]\pi/2[/itex]). But [itex]-\sqrt{3}+ i[/itex] with negative x value and positive y is in the second quadrant: the correct angle is [itex]\pi- \pi/6= 6\pi/6- \pi/6= 5\pi/6[/itex].

    And, again, for polar form you will need the modulus, [itex]\sqrt{(-\sqrt{3})^2+1^2}=\sqrt{3+ 1}= \sqrt{4}= 2[/itex].
    [/quote]


    ----





    please help me >>[/QUOTE]
     
  5. Apr 19, 2010 #4

    ----


    I have a different view of this

    let z = 1-i be a complex number first

    then we find the angle alpha[tex]\alpha = tan^{-1}(-1) = -45 = \frac{-\pi}{4}[/tex]

    since z = 1-i lies within the second quadrant

    so [tex]180--45 = 225[/tex] degrees. Which in Radians are [tex]\theta = \frac{5\pi}{4
    }[/tex]
     
  6. Apr 19, 2010 #5
    Now i understand it

    thanks


    If posiible i want all angle with degrees and radians

    for exaple : 45 = π/4
    30 = π/6

    help me
     
  7. Apr 19, 2010 #6
    You are welcome r-soy :)
     
  8. Apr 19, 2010 #7

    Mark44

    Staff: Mentor

    No, z is in the fourth quadrant.
    As a positive angle it would be 7pi/4.
     
  9. Apr 19, 2010 #8
    If posiible i want all angle with degrees and radians

    for exaple : 45 = π/4
    30 = π/6

    help me
     
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