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Complex numbers problem

  1. Mar 23, 2016 #1
    1. The problem statement, all variables and given/known data
    The complex number ##u## is defined by ## u= 6-3i/1+2i##
    i) Showing all your working find the modulus of u and show that the argument is ## -1/2π##
    ii) For the complex number Z satisfying ##arg(Z-u)= 1/4π##, find the least possible value of mod | Z |
    iii) For complex number Z, satisfying mod | Z-(1+i)u| = 1 find the greatest possible value of | Z |


    2. Relevant equations


    3. The attempt at a solution
    i) I have no problem with this one
    ##6-3i/1+2i ×1-2i/1-2i = -3i## next to get argument we shall have ## 0-3i## where sin^-1## (-3/3)=-1##
    it follows that ## ∅= -90^0 ## which is equal to ##-1/2π## which is correct answer as per marking scheme

    ii) i have a problem here, all the same my attempt
    ##Z- (6-3i/1+2i)##
    = sin^-1 ##(1/√2) ##
    this is from sin^-1 ##(1/√2) ## = ##1/4π##
    ##Z- (6-3i/1+2i)##=##1+i##
    ##Z##=##1+i+(6-3i/1+2i)##
    =##(5/1+2i)##
    and
    ##5/1+2i##
    =##1-2i##
    and
    ##|1-2i|=√5##

    This is my second attempt
    arg ##Z+(1/2π)## =##1/4π##................................................

    the correct answer to this problem is
    is ##3/2√2## kindly assist
     
    Last edited: Mar 23, 2016
  2. jcsd
  3. Mar 23, 2016 #2

    PeroK

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    First, you can't write ##1/2\pi## and expect us to know whether you mean ##1/(2\pi)## or ##\pi/2##.

    In your first step, you show that ##u = -3i##, but you don't use this simpler expression in what follows.

    I would try a geometric approach to (ii) and (iii). Can you see what these equations involving ##u## imply?
     
  4. Mar 23, 2016 #3

    BvU

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    PLease use brackets. It isn't clear whether you mean ## u= 6-3i/1+2i = 6-3i +2i##,
    ## u= 6-(3i/1+2i) ## or ##
    u= (6-3i)/(1+2i)##. I assume this last one.

    Another way is to use \frac { } { } or \over in ##\TeX##.
     
    Last edited: Mar 23, 2016
  5. Mar 23, 2016 #4
    thanks for the correction...the original problem is ## u=(6-3i)/(1+2i)## i have managed to solve both problem as follows:
    arg ##(z-u)= (1/4)π##
    ##⇒z=u , z=-3i ##now i managed to sketch this on an argand diagram....and this makes an angle of ∅= (1/4)π with the real axis. and this line z= -(1/4)π with the real axis gives us z
    z=3-3i
    |z|= √(3^2+(-3)^2)= 3√2 ......the least possible value of |z|= (3/2)√2.................this z line intersects with line from (0,0) at 90^0

    for part iii)
    locus is a circle with centre (3,-3) and radius r=1 the greatest possible value of |z| is the distance from the point (0,0) to furthest point on the circle and that is 3√2+1
     
    Last edited by a moderator: Mar 23, 2016
  6. Mar 23, 2016 #5

    PeroK

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    You perhaps need to work on explaining how you get your answers. It looks like you did exactly the right thing for parts ii) and iii) but your working isn't clear.

    You don't have to post anything more here since you have got the answers, but perhaps something to think about for future problems.
     
    Last edited by a moderator: Mar 23, 2016
  7. Mar 24, 2016 #6
    thanks for your reply though i wouldn't mind being given an alternative approach. Mathematics is'nt about answers only, it is also about understanding different approaches in solving problems, i would appreciate if there is a different approach.
    for part ii) The least distance of |z| will intersect the line from point ## (0,0)## at ##90## degrees. otherwise thank you.
     
  8. Mar 24, 2016 #7

    BvU

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    Well, as an alternative, more algebraic approach: your part ii) asks for a ##Z## for which ##\arg \left (Z-u \right) = {\pi\over 4}##.
    The argument of a complex number ##\ \alpha = x + iy \ ##is the angle ##\ \phi\ ## for which ##\ \phi = \arctan \left ( {y\over x}\right ) \ ## if ##\ x > 0 \ ##, so in your case you are looking for $$\arg \left ( Z + 3i \right ) = 1 \ \ \Leftrightarrow \ \ \operatorname{Re} \left ( Z + 3i \right ) = \operatorname{Im} \left ( Z + 3i \right ) \ \ \rm ! $$
    and with ##\ \operatorname{Re} ({\bf \alpha} + \beta) = \operatorname{Re} \alpha + \operatorname{Re}\beta \ \
    ## and ##\ \operatorname{Im} ({\bf \alpha} + \beta) = \operatorname{Im} \alpha + \operatorname{Im}\beta \ \ ## you easily find $$
    \ \ \Leftrightarrow \ \ \operatorname{Re} Z = \operatorname{Im} Z + 3 $$
    (## \& \ \operatorname{Re} Z > 0 \ ##) for which ##\ |Z|\ ## (or, easier, ##\ Z^*Z\ ##) is quickly minimized.

    (but I admit I keep a mental image of the Argand diagram in mind)
     
  9. Mar 24, 2016 #8
    Thanks Bvu
     
  10. Mar 24, 2016 #9

    Ray Vickson

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    Since ##u = -3i##, the complex number ##z-u = z + 3i## is located 3 units vertically above the point ##z = x + iy##. Thus, ##z## is located 3 units vertically below the line ##y = x##, so will be on the line ##y = x - 3##. You want to find the point ##P## on that line that is closest to the origin ##(0,0)##., and that will be where the line ##y = -x## intersects the line ##y = x-3##. Do you see why?
     
  11. Mar 28, 2016 #10
    I had already seen that bro. Thanks a lot. It infact intersects this line at 90 degrees.
     
    Last edited: Mar 29, 2016
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