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Complex Analysis

  1. Feb 8, 2013 #1
    1. The problem statement, all variables and given/known data

    1- Find the two square roots of the complex number z=3+4i.

    2a- Solve in ℂ the equations: (E): 4z^2-10iz-7-i=0

    b- Let a and b be solutions to (E) such that: Re(a)<0 and the two points A and B plots/pictures of a and b. Show that b/a=1-i. Conclude that AOB is an equilateral triangle.

    3. The attempt at a solution

    1- After solving (p+qi)^2=3+4i i found that the solutions were either 2+i or -2-i.

    2-a For the complex equation i found two complex roots: z1=(-3+6i)/8 and z2=(3+14i)/8.

    b- So i took the two solutions that i found from the previous question and chose a=z1 and b=z2 and after computing i got a whole different answer. Is my work correct, if not some help would be very much appreciated.
  2. jcsd
  3. Feb 8, 2013 #2


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    I get a different result. Pls post your working.
  4. Feb 8, 2013 #3


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    If z= (-3+ 6i)/8 then [itex]z^2= [(9- 36)- 2(18i)]/64= -27/64- (9/8)i[/itex] so [itex]4z^2- 10iz- 7- i= -27/4- (9/2)i+ (15/4)i+ 15/2- 7- i= (-27/4+ 15/2- 7)+ (15/4- 9/2- 1)i= (-27+ 30- 28)/4+ (15/4- 18/4- 4/4)i= -25/4- (7/4)i, NOT 0.

  5. Feb 8, 2013 #4
    Sorry I was wrong on the roots of the equation they are correct now i got:

    z1=(1/2)+(3i/2) and z2=(-1/2)+i and that certainly gives 1-i when you take z2=a and z1=b.

    But the conclusion i can't quite fathom, any help with that please.
  6. Feb 8, 2013 #5


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    What will the third side look like as a complex number (in terms of a and b)? What will its ratios be to the other two?
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