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: Complex number

  1. Aug 17, 2010 #1
    1. The problem statement, all variables and given/known data

    Given that arg(z/(1-i))=pi/2, find the argument of z.

    Sketch , in the Argand diagram , the set of points representing z and the point representing the complex number w=-4+3i. Hence deduce the least value of |z+4-3i|

    2. Relevant equations



    3. The attempt at a solution

    Let z=a+bi

    Rationalising z/(1-i)=(a-b)/2+(a+b)/2 i

    negative pi/2 suggest that the point is on the negative y-axis.

    tan pi/2 = [(a-b)/2]/[(a+b)/2]

    hence a=b

    And since (a+b)/2 is negative, a<0 , b<0

    so z=a+ai or b+bi

    (a,a) is in the third quadrant, and the argument of z is -3pi/4

    I have no problem in sketching. The problem i have is with the last part.

    Any pointers?
     
  2. jcsd
  3. Aug 17, 2010 #2

    Mentallic

    User Avatar
    Homework Helper

    So which is it, [itex]\pi/2[/itex] or [itex]-\pi/2[/itex]?

    You can solve the first faster and easier by using [tex]arg\left(\frac{a}{b}\right)=arg(a)-arg(b)[/tex]

    The least value of [tex]|z+4-3i|[/tex] is saying what point z where [tex]arg(z)=\frac{-3\pi}{4}[/tex] thus z=a(1+i), a<0 is closest to the point (-4+3i)? If you think about this in cartesian coordinates, z lies on the line y=x where x<0. Now all you have to do is find out the closest distance between the point (-4,3) and the line y=x.
    The problem could even be simplified using logic since we know the line connecting the shortest distance between z and the point is at right angles so the gradient of the line is -1.
     
  4. Aug 17, 2010 #3
    thanks Mentallic.
     
  5. Aug 17, 2010 #4

    Mentallic

    User Avatar
    Homework Helper

    It sure does :smile:

    If you want to prove it, let [tex]x=r cis\theta[/tex] and [tex]y=Rcis\phi[/tex] and then rationalize x/y and use some properties of trigonometry to simplify it down.
     
  6. Aug 18, 2010 #5

    HallsofIvy

    User Avatar
    Science Advisor

    It is precisely a "property of logarithms" because the argument of a complex number is a property of exponentials.

    Write complex number a, with magnitude [itex]r_1[/itex] and argument [itex]\theta_1[/itex], in the form [itex]r_1e^{i\theta_1}[/itex] and the complex number b, with magnitude [itex]r_2[/itex] and argument [itex]\theta_2[/itex], in the form [itex]r_2e^{i\theta_2}[/itex].

    Then
    [tex]\frac{a}{b}= \frac{r_1e^{i\theta_1}}{r_2e^{i\theta_2}}= \frac{r_1}{r_2}\frac{e^{i\theta_1}}{e^{i\theta_2}}= \frac{r_1}{r_2}e^{i(\theta_1- \theta_2)}}[/tex]

    That is, the magnitude of a/b is [itex]r_1/r_2[/itex] and its argument is [itex]\theta_1- \theta_2[/itex].
     
  7. Aug 18, 2010 #6
    And another student discovers the miracles of Euler's formula.
     
  8. Aug 18, 2010 #7
    Thanks Hallsofivy, and wow how long have you been on this forum?You have made 27000 posts!
     
  9. Aug 22, 2010 #8
    Is the answer for the least value sqrt(53/2) ?
     
  10. Aug 22, 2010 #9

    Mentallic

    User Avatar
    Homework Helper

    No not quite. Show us what you did.
     
  11. Aug 22, 2010 #10
    I drew a diagram, with right angled triangle WOZ. Using the perpendicular gradients,

    (a/a)((3-a)/(-4-a))=-1

    a=-1/2

    Then i proceed to find the distance between (-4,3) and (-1/2,-1/2) which gives that answer.
     
  12. Aug 22, 2010 #11

    Mentallic

    User Avatar
    Homework Helper

    Ok so you set up your question properly, you just used the distance formula incorrectly.

    [tex]d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]

    Where [tex](x_1,y_1)=(-1/2,-1/2)[/tex] and [tex](x_2,y_2)=(-4,3)[/tex]

    I can think of some other ways to solve this problem which might be of interest to you.

    So we have [tex]d=\sqrt{(x-4)^2+(y+3)^2}[/tex] to denote the distance from the point(4,-3) to a circle's circumference with that centre and radius d. We also have the equation y=x so if we plug y into the distance equation and simplify we get [tex]d=\sqrt{2x^2-2x+25}[/tex] but we want to find the minimum distance so we want the minimum value of the quadratic under the sq.root. Just use some calculus and you'll find x=1/2, which can be substituted back into the distance equation to give you the answer.
     
  13. Aug 22, 2010 #12
    thanks Mentallic, it can also be solved using trigonometry and vectors.
     
  14. Aug 22, 2010 #13

    Mentallic

    User Avatar
    Homework Helper

    Yep, and if you go back to precalculus class they would be using a specific distance formula for the distance between a point and a line :wink:

    By the way, I have no idea why this was moved to the precalc forum.
     
  15. Aug 22, 2010 #14

    HallsofIvy

    User Avatar
    Science Advisor

    I started last night and, boy, are my fingers tired!:rofl:
     
  16. Aug 22, 2010 #15

    Mentallic

    User Avatar
    Homework Helper

    Halls, an average of 10 posts per day for the passed 7 years... umm... just one question. Assuming you have to take a break once a year or so, for christmas maybe :wink: how many posts have you made on your big days? I'm suspecting you spammed the general forum like hell :tongue:
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook