1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: I want help to find tan of polar correctly

  1. Apr 19, 2010 #1

    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


    User Avatar
    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].


    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
  6. Apr 19, 2010 #5
    Now i understand it


    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


    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook