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Finding all pairs of values that satisfy complex equation

  1. Oct 24, 2013 #1

    BMW

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    1. The problem statement, all variables and given/known data
    Find all pairs of values a and b that satisfy (a + bi)2 = 48 + 14i

    2. The attempt at a solution
    I managed to solve it, but it took a while and I was wondering if there is an easier/quicker way.

    What I did was expanded (a + bi)2 into (a2 - b2) + 2abi
    From there, I can now say:
    a2 = b2 = 48
    2ab = 14

    I then got a value for b from the 2ab = 14 equation (b = 7/a) and substituted this into the a2 - b2 = 48 equation, and I got this polynomial:
    a4 - 48a2 - 49 = 0

    To solve the polynomial, I used the factor theorem and found that (a + 7) and (a - 7) are two factors, then I used synthetic division to divide the polynomial by these two factors (a + 7) and (a - 7), to get the quadratic a2 + 1 = 0

    Solving this quadratic resulted in the last two factors being (a - i) and (a + i)

    So now I have the factorised polynomial (a - 7)(a + 7)(a - i)(a + i) and can get the values for a:
    a = 7, -7, i, -i

    Now I substitute those values into the equation used earlier (a2 - b2 = 48) to get the values for b, resulting in these pairs:
    a = 7, b = ±1
    a = -7, b = ±1
    a = i, b = ±7i
    a = -i, b = ±7i

    And from there I substitute all pairs into the original equation to check their validity and remove the invalid ones.

    It's quite a procedure for a relatively simple problem, is there any quicker way?
     
  2. jcsd
  3. Oct 24, 2013 #2

    Mark44

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    Up to here, I don't think there are any shortcuts. Now, instead of what you did below, use your other equation (b = 7/a) to find the value of b for each value of a.

    It seems that you are going backwards in using the equation a2 - b2 = 48. You had already rewritten it as a4 - 48a2 - 49 = 0, and that's how you got your a values.

    If you take those 4 values of a and calculate b as 7/a, you get the 4 corresponding b values.
     
  4. Oct 24, 2013 #3

    BMW

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    Thank you, that makes it a bit easier.

    In the answers for that question, they go from a2 - b2 = 48 to the solutions. It seems they are skipping a lot, so that's why I thought there must be an easier way.
     
  5. Oct 24, 2013 #4

    haruspex

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    Looking at the magnitudes, a2+b2 = √(482+142)=50. Combine that with your a2-b2 = 48.
     
  6. Oct 24, 2013 #5

    BMW

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    Nice idea, but does that get the complex solutions? I tried it and it didn't seem to...
     
  7. Oct 25, 2013 #6

    haruspex

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    In the OP you show too many solutions. By definition, a and b are real. The second pair of solutions you show is actually the same as the first pair, but with a and b sort of swapped. More precisely, they swap a with ib, etc.
     
  8. Oct 25, 2013 #7

    Mentallic

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    That's a lot of unnecessary work. Notice that if you let [itex]c=a^2[/itex] then

    [tex]a^4-48a^2-49=0[/tex]
    becomes
    [tex]c^2-48c-49=0[/tex]

    Which is a quadratic in c, and factorizing that gives us

    [tex](c-49)(c+1)=0[/tex]

    Hence [itex]c=-1,49[/itex] and now converting back to a, [itex]a^2=-1,49[/itex] then [itex]a=\pm i, \pm 7[/itex]

    But you can of course skip the substitution part as so:

    [tex]a^4-48a^2-49=0[/tex]

    [tex](a^2-49)(a^2+1)=0[/tex]

    [tex]a^2=-1,49[/tex]

    And as haruspex said, a and b are assumed to be real, so you can toss out the complex values for a, leaving you with [itex]a=\pm7[/itex]

    From here, you already had that b=7/a, so it's easy to see that if [itex]a=\pm7[/itex] then [itex]b=7/\pm7=\pm1[/itex]
     
  9. Oct 25, 2013 #8

    BMW

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    Why are a and b assumed to be real? It doesn't say anything about that in the question, and if you substitute the imaginary solutions into the original equation, it works. (e.g. the solution a = i, b = -7i works)
     
    Last edited: Oct 25, 2013
  10. Oct 25, 2013 #9

    Mentallic

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    Because when you let [itex]z=a+ib[/itex] then you assume that a,b are real. Of course, if a,b are complex values, then you can still simplify the expression into the form z=x+iy where x,y are real.

    Yes, it works, but it gives you the same results that you would get when only considering [itex]a=\pm7[/itex]. a=i, b=-7i gives us

    [tex]a+ib = i+i(-7i) = 7+i[/tex]

    Which is the same as the solution a=7, b=1.

    It's just easier to assume they're real variables so that you can toss out 2 of the 4 solutions to the quartic.
     
  11. Oct 25, 2013 #10

    haruspex

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    I would phrase it a little more strongly: you are defining a and b to be real. If you do not specify that then they are not fully defined. E.g. you could have a = 7+i, b=0. There becomes an infinity of (equivalent) pairs of solutions, not just the two pairs in the OP.
     
  12. Oct 25, 2013 #11

    Mentallic

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    Ahh yes, you're right!
     
  13. Oct 26, 2013 #12

    BMW

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    But I'm not letting z = a + ib. I am simply trying to find all the pairs of a and b that satisfy the equation (a + bi)2 = 48 + 14i
    The way I think of it, is that the expression in brackets (a + bi) is not a complex number, its a number (real, imaginary, or complex) added to another number (real, imaginary, or complex) multiplied by i. Nothing in the question said that the expression in the brackets was a complex number, so we cannot assume that a and b are real.

    The answer to that question in the textbook has the imaginary solutions as well, why toss out solutions only to get the answer wrong? The imaginary values do satisfy the equation, so they should be included in the answer. That is what the question is asking.

    Well yes the solution a = 7 + i, b = 0 does work... now I'm really confused. The answers in my textbook have the imaginary solutions for a and b, so they aren't necessarily real, or are the answers wrong?
     
    Last edited: Oct 26, 2013
  14. Oct 26, 2013 #13

    Mentallic

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    The form that a complex number takes is z=a+ib where a,b are real numbers. If you were asked to solve [itex]z^2=48+14i[/itex] where z is a complex number, then the first thing you would do would be to make the substitution z=a+ib.

    Remember that a complex number is any number of the form a+ib, where a,b are real, which also means that b=0 is still a valid complex number. So any real number is also a complex number, and even the natural numbers are a subset of the complex numbers. It's just like a square is a special kind of rectangle, and both are a special kind of parallelogram, which itself is a special kind of quadrilateral. The complex numbers can be considered as being the quadrilateral.


    You aren't getting the answer wrong if you discard the imaginary solutions, you're merely duplicating the solutions you got when only using [itex]a=\pm7[/itex].



    The answers in your book shouldn't include the imaginary solutions for exactly the reason that haruspex stated.
    a,b must be real numbers.

    The values for a given by the quartic are the real values of a, and the purely imaginary values of a (complex numbers with real part equal to 0, or in other words, of the form a+ib where a=0).
    If you wanted to let a,b be any complex numbers, then you'd let a=c+id and b=x+iy where c,d,x,y are strictly real. So then

    [tex]a+ib = (c+id)+i(x+iy)[/tex]

    [tex]=c+id+ix-y[/tex]

    [tex]=(c-y)+i(d+x)[/tex]

    and then if you solved for these values, you'd find that you'd have an infinite number of solutions for a and b, and specifically, there would be only two real solutions for a (these values you already know) and two purely imaginary solutions for a. The rest would all be complex with [itex]a,b\neq 0[/itex].
     
  15. Oct 26, 2013 #14

    BMW

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    Ok so the textbook is wrong in having the imaginary solutions?
     
  16. Oct 26, 2013 #15

    Mentallic

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    It's not a wrong solution, but it's a duplicate solution so it shouldn't have been included.

    If you're asked to solve for x in [itex]\sin(x)=1[/itex] and the text book gives the solutions [itex]x=\pi/2, 5\pi/2[/itex] it has given correct answers, but its second solution is a period of [itex]2\pi[/itex] larger than the first so in a sense it's a duplicated solution. It has neither given the principal angle [itex]x \in [0,2\pi)[/itex], but it also hasn't given all infinite solutions. If the question stated that [itex]x \in [0,2\pi)[/itex] and it had given a value of x outside of this domain, then yes, it's wrong. In the same sense, if the question specifically stated that [itex]a,b\in \mathbb{R}[/itex] then the imaginary solutions given for a in the textbook would be incorrect, but since it doesn't state that, it's not strictly incorrect.

    Most importantly, you're "more correct" if you only choose [itex]a=\pm7[/itex] and ignore the imaginary solutions.
     
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