# Solving a system of Non-linear Equations with Complex numbers

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Hi all,

I would like to solve the following system of 3 non-linear equations in 3 unknowns :

function F= myfun(x)

F=[ x(1) + x(2) - 10000;

x(1)*exp(-1i*x(3)*5) + x(2)*exp(1i*x(3)*5) - 12000;

x(1)*exp(-1i*x(3)*10) + x(2)*exp(1i*x(3)*10) - 8000 ];

The fsolve command does not solve complex number roots. And splitting it up into real and imaginary also does not seem to work (I took the help of Euler's formulae to convert to cosine and sine).

Please help me out!

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### Answers (4)

John D'Errico
on 6 Apr 2017

Edited: John D'Errico
on 6 Apr 2017

It appears there are two solutions. They are best found using solve, so a symbolic solution.

syms x1 x2 x3

eq1 = x1 + x2 - 10000;

eq2 = x1*exp(-1i*x3*5) + x2*exp(1i*x3*5) - 12000;

eq3 = x1*exp(-1i*x3*10) + x2*exp(1i*x3*10) - 8000;

result = solve(eq1,eq2,eq3,x1,x2,x3)

result =

struct with fields:

x1: [2×1 sym]

x2: [2×1 sym]

x3: [2×1 sym]

result.x1

ans =

5000 - (7^(1/2)*9000i)/7

(7^(1/2)*9000i)/7 + 5000

result.x2

ans =

(7^(1/2)*9000i)/7 + 5000

5000 - (7^(1/2)*9000i)/7

result.x3

ans =

-(log(3/4 - (7^(1/2)*1i)/4)*1i)/5

-(log((7^(1/2)*1i)/4 + 3/4)*1i)/5

vpa(result.x1)

ans =

5000.0 - 3401.6802570830450449306488261076i

5000.0 + 3401.6802570830450449306488261076i

vpa(result.x2)

ans =

5000.0 + 3401.6802570830450449306488261076i

5000.0 - 3401.6802570830450449306488261076i

vpa(result.x3)

ans =

-0.14454684956268312223567547052827

0.14454684956268312223567547052827

As you can see by simple tests to verify this is a solution...

result.x1 + result.x2

ans =

10000

10000

vpa(result.x1.*exp(-1i*result.x3*5) + result.x2.*exp(1i*result.x3*5) - 12000)

ans =

0

0

##### 0 Comments

Star Strider
on 1 Jul 2014

‘The fsolve command does not solve complex number roots.’

It does for me!

Your function (as an anonymous function):

myfun = @(x) [ x(1) + x(2) - 10000;

x(1)*exp(-1i*x(3)*5) + x(2)*exp(1i*x(3)*5) - 12000;

x(1)*exp(-1i*x(3)*10) + x(2)*exp(1i*x(3)*10) - 8000 ];

X = fsolve(myfun, [1; 1; 1])

produced:

X =

2.3160e+000 - 33.3155e-003i

2.3160e+000 + 33.3155e-003i

1.2720e+000 + i

It wanted me to let it do more function evaluations, but I’ll leave that to you. (I’m running R2014a, but I doubt that makes a difference.)

##### 2 Comments

Rubron
on 6 Apr 2017

John D'Errico
on 6 Apr 2017

Edited: John D'Errico
on 6 Apr 2017

Sorry. lsqnonlin will definitely not work happily with complex variables, which explains your failure to get it working.

Hmm. I did just notice that the solution Star proposed does not actually solve the system of equations. For example, the sum x(1)+x(2) does not yield 10000. In fact, that sum is not even close to 10000. So I'm not too sure that fsolve is terribly happy at solving complex problems either.

Alex Sha
on 18 Nov 2022

There are much more solutions else:

x1: 5000+3401.68025708298i

x2: 5000-3401.68025708301i

x3: -3.62536433474507+0i

x1: 5000+3401.68025708298i

x2: 5000-3401.68025708301i

x3: 1.11209021187323+0i

x1: 5000+3401.68025708298i

x2: 5000-3401.68025708301i

x3: 2.36872727330915+0i

##### 0 Comments

David Goodmanson
on 18 Nov 2022

Edited: David Goodmanson
on 18 Nov 2022

Hellol Aravindhan,

not solved by Matlab, but a bit of algebra shows that

with

x(1) = 2000*a

x(2) = 2000*b

x(3) = theta/5

then

theta = acos(3/4);

a = (5/2 +9*i/(2*sqrt(7)));

b = (5/2 -9*i/(2*sqrt(7)));

format long g

theta = acos(3/4)

x = [2000*a 2000*b theta/5]'

theta = 0.722734247813416

x = 5000 - 3401.68025708305i

5000 + 3401.68025708305i

0.144546849562683 + 0i

There are an infinite number of solutions to theta = acos(3/4). First of all there is the 2pi ambiguity, so theta = .7227 + 2*pi*n is a set of solutions. Then the negative angle, -.7227 (with its 2pi ambiguity) is a set of solutions as well. But note that the equations are symmetric under theta --> -theta, a<-->b

so the -.7227 set is the same thing with a and b swapped. Consequently there are only two possibilities for a and b.

##### 0 Comments

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