How to solve two equations involving complex variables?

[shaiqbashir]

how to solve two equations involving complex variables??

Hi guys!

well! i want to know how can u solve the following two equations simultaenously in order to find out Ix and Iy:


(3+j4)Ix - j4Iy=10-------------(1)
(2-j4)Ix +j2Iy=0---------------(2)

please tell me the best methods to solve these two equations. How can i solve them with my scientific calculator?

Please tell me both ways:

solving manually

and solving through scientific calculator.

I shall be thankful to u for this act of kindness.

Take carez!

Good Bye!

 

[HallsofIvy]

What do "j4" and "J2" mean? I assume j is your engineer's way of writing the imaginary unit that I would call i, but what are the 4 and 2 after them. If they are intended to mean powers, i^4 and i^2, then i²= -1 and i⁴= 1, of course.

 

[sonutulsiani]

HEy it's the way we write in circuits. I am looking for the same thing, how do you solve equations involving complex numbers by hand? My calculator doesn't do it.

So according to his equation, it means:

(3+4i) Ix - 4i Iy=10-------------(1)
(2-4i) Ix +2i Iy=0---------------(2)

And it's not a square of i. Please explain. Thanks.

 

[sonutulsiani]

We use j instead of i in circuits. And 3i is written as j3. It means the same.

 

[NaturePaper]

@sonutulsiani,
And what about that big 'I'? What is meant by Ix, Iy etc?

 

[sonutulsiani]

Ix and Iy are currents. Do you know how to solve?

 

[Hurkyl]

You posed your question as if the fact the coefficients are complex numbers with nonzero imaginary parts is relevant. Why?

 

[sonutulsiani]

You posed your question as if the fact the coefficients are complex numbers with nonzero imaginary parts is relevant. Why?

I didn't understand what you said.

 

[MustangGT]

It's just algebra, you have two equations in two unknowns. You can solve for Ix in terms of Iy and then do substitution.

The complex coefficients you have represent the reactive components in your circuit, if they don't end up canceling then your currents will be out of phase with your voltage and how far out they are will be determined by the angle you get when you change your answers to polar form.

 

[sonutulsiani]

Can't we solve it by separating the real and imaginary parts both on the left and right hand side of the equation?

 

[NaturePaper]

Ix and Iy are currents. Do you know how to solve?

I don't know about currents...all I'm asking whether they are numbers (here variables) or not. If they are just variables (I would like x instead of Ix and y instead of Iy), then it is very easy to solve. (No need to separation). I'm finding the value of Ix below and hope you will be able to follow:

(equation 1).(2i)-(equation 2). (-4i) gives

Ix=[20i]/[(3+4i).2i+(2-4i).4i]=[10]/[(3+4i)+4-8i]=10/(7-4i)=10(7+4i)/(49+16)=(14/13)+(8/13)i

Please check, I may have done any silly mistake in calculation.

Regards,

 

[MustangGT]

your answer is correct.