Real Value of x in Complex Equation?

  • Thread starter Alec
  • Start date
In summary, the problem asks to determine the value of x that makes the real part of the equation (10x-40i)/(x^2+16) equal to 1. The solution involves taking the real part of the equation and setting it equal to 1, then solving for x.
  • #1
Alec
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Homework Statement


Decide the real value of x making Re(10 / x+4i) = 1


Homework Equations


I know that you can extend the equation with x-4i.


The Attempt at a Solution


But then the equation looks like this: (10x-40i) / (x^2 + 16) = 1
I don't know how to get rid of the -40i to make it into a real equation.

Any help would be valuable!
 
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  • #2
[tex]\frac{10x-40i}{x^2+ 16}= \frac{10x}{x^2+16}-i\frac{40}{x^2+ 16}[/tex]
 
  • #3
It says the real part of (10x - 40i)/(x^2 + 16) = 10x/(x^2+16) - 40i/(x^+16). Take the real part of that and set it equal to 1, and solve for x.
 

1. What does "Re" mean in this equation?

The "Re" in this equation refers to the real part of a complex number. In other words, it is the numerical value of the complex number without the imaginary component (in this case, 4i).

2. How do I solve for x in this equation?

To solve for x in this equation, you can start by isolating the variable on one side of the equation. In this case, you can subtract 1 from both sides, then multiply both sides by x+4i. This will leave you with a quadratic equation that you can solve using the quadratic formula or by factoring.

3. Can this equation have more than one solution for x?

Yes, this equation can have multiple solutions for x. This is because the complex numbers have both a real and imaginary component, meaning there can be multiple values that satisfy the equation.

4. Can I solve this equation without using complex numbers?

No, this equation requires the use of complex numbers to solve for x. The imaginary component is essential in finding the solutions to this equation.

5. How can I check my solution?

To check your solution, you can substitute the value you found for x back into the original equation. If it satisfies the equation, then it is a valid solution. You can also use a graphing calculator to graph the equation and confirm that the value you found for x is the intersection point on the graph.

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