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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

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    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

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    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

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    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

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    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

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    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

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    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

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

    Mentallic

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    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:
     
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