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Complex numbers ->equations.

  1. Feb 18, 2014 #1
    1. The problem statement, all variables and given/known data

    Solve the equations:

    3(z-2) = 2j(2z+1)

    and

    (i-2)z-z*=3i+1

    where z* is the complex conjugate of z.

    (I am assuming z and z* are the unknowns. i and j are basically the same since they're defined as i2 = j2 = -1?)

    2. Relevant equations

    Rules for solving regular equations?

    3. The attempt at a solution

    3(z-2) = 2j(2z+1)

    3z - 6 = 4zj + 2j

    3z - 4zj = 2j + 6

    How do I proceed from here? I want to multiply every number with j to do j2 = -1, but then I'm left with

    3zj + 4z = 6j - 2 which basically gets me nowhere.

    2nd equation:

    (i-2)z-z*=3i+1

    zi - 2z - z* = 3i +1

    Same problem here. Want to multiply each number by i, but then I'm left with

    -2zi -z -z*i = 1 - 3

    I'm using the same rules as I'm using at single equations with one unknown and the same rules as two equations two unknowns in "normal" algebra. I want to proceed with z = something in my first equation and put that into equation 2 and solve.

    Any ideas, guidelines or input would be very much appreciated.
     
  2. jcsd
  3. Feb 18, 2014 #2
    Isolate z on one side of the equation. Don't worry about the j until the end (when you put the answer in standard a + jb form).



    I would personally rewrite z and z* as a + bi and a - bi, respectively, and solve for a and b. I'm sure there is a clever way to rewrite z* in terms of z and solve it in an easier way, but I like the components method. Again, don't worry about trying to get rid of the i. Almost all of your answers in Complex Analysis will involve i.
     
  4. Feb 18, 2014 #3

    SteamKing

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    Hint: let z = a+ib, and you know then that z* = a-ib. Substitute into your original equations, and remember for complex equations to be true, the real parts are equal and the complex parts are equal.
     
  5. Feb 18, 2014 #4
    Hmm. I'm a little bit confused, although I understand everything you're both saying.

    1: If I substitute z = a + ib and z* = a- ib into the original equations, should I proceed like this:

    3(z-2) = 2i(2z+1) -> 3((a+ib)-2) = 2i(2(a+ib)+1)

    then I would get 3(a+ib)-6 = 4i(a+ib)+2i

    Should I then proceed to put a+ib on one side? This was very confusing!

    I'm tempted to both divide every number in the equation by 4i, and multiply the 4i(a+ib) to get 4ai + ##4bi^2##

    Thanks so far, I really appreciate it. I feel I'm a step closer to the answer here :)
     
  6. Feb 18, 2014 #5
    I personally think converting to a + ib format for the first problem is more tedious, but you could do well with the practice, so let's stick with it, Following SteamKing's hint (real parts are equal, complex parts are equal), you want to eventually have your equation in the form ___ + ___ i = ___ + ___ i where all of your variables (a,b) are on the left side of the equation and the numbers are on the right side.

    For example, (2a + b) - bi = 1 + 3i would be in the form above. In my example, you would then equate like terms and solve the system of equations that follows. (2a+b = 1, -b = 3)

    So, to answer your question, don't treat the a + bi as a single term. Multiply everything out and group all the terms based on whether they are imaginary or real.
     
  7. Feb 18, 2014 #6
    Ok, this just made sense to me - I think. It's a little confusing as our standards here are slightly different (just in regards to what order we place the real number and the imaginary number etc - no biggie (like we use , for decimals heh)).

    I still struggle with the last part, but here's my progress now:

    First problem:

    3(z-2) = 2i(2z+1)
    3z-6=4zi+2i
    3z-4zi=6+2i
    (3-4i)z=6+2i
    z=[itex]\frac{6+2i}{3-4i}[/itex]=[itex]\frac{(6+2i)}{(3-4i)}[/itex]*[itex]\frac{(3+4i)}{(3+4i)}[/itex] = [itex]\frac{18+24i+6i+8i^2}{3^2 + 4^2}[/itex] = [itex]\frac{18-8+30i}{25}[/itex] = [itex]\frac{2}{5}[/itex]+[itex]\frac{6}{5}[/itex]i

    2nd problem:

    (i-2)z - z* = 3i+1
    zi-2z-z* = 3i +1
    z = a+ib and z* = a-ib

    i(a+ib)-2(a+ib) - (a-ib) = 3i+1
    ia + ##i^2##b-2a-2ib-a+ib = 3i +1
    ia-b-2a-2ib-a+ib=3i+1
    -3a-b+ia-ib=3i+1
    -3a-ib = 1+3i

    I know I just removed my -y and ix from the equation at my last part, but they just don'e fit in! how did I get them, and how do I get rid of them? I will go through my basic algebra here and edit my post if I spot any careless mistakes, but my mind is pretty frustrated with this now so I'm sorry if I fail to spot any obvious mistakes.

    Thanks so much for all your help so far - I feel like I'm getting very close to the answer now :)
     
    Last edited: Feb 18, 2014
  8. Feb 18, 2014 #7
    All the work looks good to me, I didn't spot any mistakes. Problem 1 looks correct to me. As for problem 2, you are right at the end of the problem. The final step is to equate like terms and solve for a and b. Equate the two real parts and equate the two imaginary parts (leaving off the i). Restating my suggestion from my post above:

    Once you find your a and b, use them to find z and you will solve the problem. Good work so far!
     
  9. Feb 18, 2014 #8
    I got this good feeling when I finally saw the first problem, but now it's back to slightly frustrated. I'll post this attempt, though, before I feel like an idiot again heh.

    So I solve for a and b, a being the real part, b being the imaginary part. And I set them both to equal 0, for some reason.

    -3a-b+ia-ib=3i+1

    gives:

    -3a-b+1 = 0 (real part) --> a=(-b+1)/3

    a-b-3 = 0 (imaginary part) --> b = -4 --> a = 5/3

    z =5/3 - 4i

    I fear I'm confusing my notes and your help - I honestly have absolutely no idea if what I've done here makes any sense at all.
     
  10. Feb 18, 2014 #9
    Unfortunately, not correct. You don't want to set the equations equal to 0. You have -3a-b+ia-ib=3i+1. I'll add some parentheses for you:

    (-3a-b) + (a-b)i= 1 + 3i

    ##\implies##

    -3a - b = 1, (equating the real parts)
    a - b = 3, (equating the imaginary parts and dropping the i)

    Do you see how I got that? Now solve for a and b.
     
  11. Feb 18, 2014 #10
    Ah, of course! When I wrote that I didn't know why I set them equal to 0, I meant it! Now I see where to get the numbers, and it couldn't be more obvious.

    I feel I've done the right thing now, but I wouldn't be too surprised if I've messed up again, but here it goes:

    (-3a-b) + (a-b)i= 1 + 3i

    ##\implies##

    -3a - b = 1
    a - b = 3

    -3a-b=1 ##\implies## -3a=b+1 ##\implies## a=[itex]\frac{-b-1}{3}[/itex]

    [itex]\frac{-b-1}{3}[/itex] -b = 3

    (-b-1)-3b=9
    -4b=10
    b=[itex]\frac{5}{2}[/itex]

    a=[itex]\frac{-\frac{5}{2}-1}{3}[/itex]= -[itex]\frac{7}{6}[/itex]

    z=-[itex]\frac{7}{6}[/itex] + [itex]\frac{5}{2}[/itex]i
     
  12. Feb 18, 2014 #11
    Looks good except the b is negative (when you solved for a you had it negative but not when you solved for it and in the final solution). Also, you probably picked the most complicated way of solving the system, wouldn't it be better to start with the second equation, as opposed to solving for a in the first? As long as you get the correct answer, the approach doesn't matter. In the future just look for the easier way to solve systems (avoid a lot of fractions).

    The final answer looks good though (except for the sign of b)! If you need more practice, try doing the first problem with this method if you want.
     
  13. Feb 18, 2014 #12
    I did the same mistake on paper - not sure why. Careless I guess.

    In the course I'm taking, we go weeks or a couple of months between working with these methods, so I often forget the small rules to make the job easier, but I definitely remember being told on campus to check which equation looks easier to swap - just like you said.

    I will keep it in mind, and definitely check back to this topic whenever I'm stuck with anything similar again.

    Thank you so much for your help, I really appreciate it.
     
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