Register to reply 
On solving for a variable in the magnitude of a complex number 
Share this thread: 
#1
Dec2212, 07:36 PM

P: 41

All,
I've been beating my head up on this equation, so I thought I'd try here to see if someone could help me. In the equation (sqrt{32}(2xd))^4 = (d/2  1/2 )^2 + sqrt{(1+d)^2 + d/4}/4 Where the right hand side is my attempt to solve for d out of the magnitude z of a complex number. You can see by looking at the right hand side (if you look carefully) how the right side used to be expressed in the form z=r=sqrt{x^2+y^2}, where z = x + yi, the magnitude (modulus) of a complex number. I got kind of tangled up with this, but I've left it where I got stuck, as it looks like an algebraic solution is possible (a polynomial of sorts), but I see I am missing some information on how to proceed and I wanted to see if someone could help me where I left off. And no, this is not a homework question. In short, my question is, with so many power and sqrt's, how does one begin to solve for d, as the only other variable is x, which I already have a solution for. Is this a 3rd or 4th degree polynomial, or is it simpler? Or is it more complicated still? Any help and pointing me in the right direction? I've tried many attempts, but feel there might be a simple method I am just over looking...or perhaps some aspect of the mathematics that I have not been involved with. Thanks, Jeff 


#2
Dec2212, 09:38 PM

Sci Advisor
P: 3,297

I think you'll get better advice if you state the original problem that you're trying to solve rather than starting midway in your attempt to solve it.



#3
Dec2212, 10:24 PM

P: 41

I know you're right. I Left this problem on my computer at work and wrote it from home. But let me try this, it won't amount to my solution, but maybe it will help me understand so I can get to the solution.
Let z be calculable from d such that z becomes an argument for a 2nd degree polynomial, and d is always negative so that z is always complex when taking its square root in the quadratic equation, the modulus of the quadratic solution is equal to some multiplicative constant of xd, how does one solve for d in...say...(which is pretty close as memory serves to my original problem) (xd)^2 = ((1d) +/ sqrt((1d)^2+d/4))/2 = z =((1d)/2)^2 +/ (i sqrt((1d)^2+d/4)/2)^2 such that z = sqrt{a^2+b^2} Such that a = ((1d)/2 b = i sqrt((1d)^2+d/4)/2) Not sure if that helps. Both are two different equations involving the same symbolized constants and the same situation, but I'm probably off from my original just but a sign. what happens when solving for these is that the powers starts to rise for any of my attempted solutions, as we have taken the negatives of d in so many cases (1d). Take the first step simply considering one of the solutions for z... (xd)^2 =((1d)/2)^2 + (i sqrt((1d)^2+d/4)/2)^2 or a^2=b^2+ic^2 But we want some constant value out of a, b, and c. This is pretty much the problem. How do I get d out of a, b and c? Thanks, Jeff 


#4
Dec2212, 10:34 PM

P: 41

On solving for a variable in the magnitude of a complex number
Would it help or confuse the issue if I said that z becomes the elements of a Hermitian matrix where the two solutions for z are complex conjugates?



#5
Dec2212, 10:38 PM

P: 41

I forgot my square root above in
=((1d)/2)^2 +/ (i sqrt((1d)^2+d/4)/2)^2 Should be =sqrt(((1d)/2)^2 +/ (i sqrt((1d)^2+d/4)/2)^2) 


#6
Dec2212, 10:42 PM

P: 41

Also forgot it here...
a^2=b^2+ic^2 Should be a^2=Sqrt{b^2+ic^2} 


#7
Dec2312, 12:38 AM

Sci Advisor
P: 3,297




#8
Dec2312, 06:56 AM

P: 41

It can be expressed in terms of two Hermitian Matrices, one 2X2 if you consider z and the d the limits of two parameterized constants z(t), d(t), and the other an infinite Hermitian matrix including all points in z(t) and d(t). I typically work with the limits, thus 2X2 for solutions involving the two. Because z is always produces a complex number for any real or complex argument to limits of its domain, the 2X2 limit matrix is useful. I've been working with this for over two years (not the current problem...just this particular relationship between d and z).



#9
Dec2312, 07:30 AM

P: 41

Iif there is no straight forward solution to the above alone as it is expressed, then I should explain I am working with two matrices, one proportionate to another. d is the divisor of the differentiation operator of another matrix, a matrix of a probability distribution. Hope those facts of the matrix help a bit. Happy to provide more info if need be. Thanks.



#10
Dec2312, 12:33 PM

P: 41

Stephen,
I think I found the fundamental problem with this equation and I have been able to simplify it. Considering a, b and c (ab)^4  b^2 = sqrt{b} = c There is only one possible argument for a that will ever provide a solution: a = lim A If A is a function, then (lim (A)b)^4  b^2 = sqrt{b} = c, then c has only two arguments b that permit its solution c. Let b = 0, b=1 and plug them one at a time into (ab)^4  b^2 = sqrt{b} = c and take A to infinity, the left hand side converges to the right (rather quickly for many arguments). Thus, they are not equal...they are asymptotically equivalent, considering A a function (which in my original example it is). Because of the rapid convergence of c, just about any decimal precision used might at first make it appear equivalent. But they're not exactly. The rest of my original equation then is simply a variation of the following (lim(a)b)^4  b^2 = sqrt{b} = c Does this make sense? Jeff 


#11
Dec2312, 01:04 PM

Sci Advisor
P: 3,297




#12
Dec2412, 10:40 AM

P: 41

Whatever, Stephen. Good luck. And have a good holiday.



Register to reply 
Related Discussions  
Complex Number Express in magnitude/phase form  Calculus & Beyond Homework  1  
Solving the Equation for a Complex Number  Calculus & Beyond Homework  5  
Magnitude of this complex number  Precalculus Mathematics Homework  1  
Magnitude of a complex number  Precalculus Mathematics Homework  4  
What is the magnitude and angle of this complex number?  Introductory Physics Homework  4 