- #1
Wardlaw
- 27
- 0
Homework Statement
(x+iy)=[1/x-iy]+2
jegues said:I'm confused at what your trying to do here? Could you state the question you were given for us? Your attempt at the solution?
jegues said:Hmmm... I'm not sure if this right but here's what I did.
I first grouped all the real and imaginary parts together on the LHS of the equation giving me the following
[ (x^2-2x-1) / x ] + 2iy = 0
You can use quadratic formula to solve the roots of the numerator in the real part and as for the imaginary parts you can simply set y = 0.
So, x = 1 +or- sqr(2) and y = 0 should satisfy the equality if I'm not mistaken.
I'm not too sure about this so maybe have someone reapprove my thoughts.
Hope this helps,
jegues said:In the numerator your have a quadratic of the form Ax^2-Bx-C, so to find its roots just simply apply the quadratic formula.
So, x = ( -B +or- sqr[ (B)^2 - 4(A)(C) ] ) / 2A
jegues said:Mentallic has the right solution, you didn't use brackets properly in your OP so saw it as 1/x - iy, not the proper form,
1/(x-iy)
jegues said:Mentallic has the right solution, you didn't use brackets properly in your OP so saw it as 1/x - iy, not the proper form,
1/(x-iy)
jegues said:Hmmm... I'm not sure if this right but here's what I did.
I first grouped all the real and imaginary parts together on the LHS of the equation giving me the following
[ (x^2-2x-1) / x ] + 2iy = 0
You can use quadratic formula to solve the roots of the numerator in the real part and as for the imaginary parts you can simply set y = 0.
So, x = 1 +or- sqr(2) and y = 0 should satisfy the equality if I'm not mistaken.
I'm not too sure about this so maybe have someone reapprove my thoughts.
Hope this helps,
are exactly the same. haha that would've been bad for OP to find that the solutions you've shown are correct, so then he goes ahead and uses that method for every other question of this type.
But still, you should've realized what the OP was trying to show by the way the brackets were placed
Complex numbers are numbers that have both a real part and an imaginary part. They are represented in the form a + bi, where a and b are real numbers and i is the imaginary unit, equal to the square root of -1.
To solve a complex number equation for x and y, you can use a variety of methods such as substitution, elimination, or graphing. However, the most common and efficient method is to use the quadratic formula.
Yes, complex number equations can have multiple solutions for x and y. This is because there are infinite solutions to the square root of -1, which is represented by i in complex numbers. Therefore, when solving these equations, you may end up with multiple values for x and y.
Real solutions are solutions that are represented by real numbers, while imaginary solutions are represented by the imaginary unit i, which is equal to the square root of -1. In complex number equations, both real and imaginary solutions may be present.
To check if your solution is correct, you can plug the values of x and y back into the original equation and see if it satisfies the equation. If it does, then your solution is correct. Additionally, you can also use a graphing calculator to visually confirm your solution.