Did you try setting ##k= \pm 2## in the original number to see whether you get a real number?squenshl said:Homework Statement
Find all possible values of ##k## that make ##u = \frac{k+4i}{1+ki}## a purely real number.
Homework Equations
The Attempt at a Solution
I calculated the complex conjugate which was ##\frac{5k}{k^2+1} + \frac{4-k^2}{k^2+1}i##. So to prove this do I just solve ##\frac{4-k^2}{k^2+1}## = 0 for ##k##?
In this case ##k = \pm 2##. Thanks.
It does look good!squenshl said:Yes to the first and yes to the second (conjugate of the denominator though) and it looks all good!
In this context, "purely real" refers to a mathematical expression or equation that has a real number solution. This means that the solution does not involve any imaginary numbers, which are numbers that involve the square root of a negative number.
In this equation, "u" represents a variable that can take on any real value. The goal is to find values of "k" that will result in a purely real solution for "u".
Finding the real values of "k" is important because it ensures that the solution for "u" is a real number and does not involve any imaginary numbers. This is important in many scientific and mathematical applications where real numbers are necessary.
To determine the real values of "k" for this equation, you will need to use algebraic methods and solve for "k". This may involve simplifying the equation, factoring, or using the quadratic formula. Once you have found the real values of "k", you can plug them back into the equation to find the corresponding values of "u".
Yes, there are some general rules and guidelines that can help you find the real values of "k" for this equation. For example, if the equation has a square root, "k" must be greater than or equal to 0 to ensure a real solution. Additionally, if the equation has a fraction, the denominator cannot be equal to 0. It is also important to check your solution to make sure it satisfies the original equation.