Find Real Values of k for Purely Real ##u##

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Homework Help Overview

The problem involves finding values of ##k## that make the expression ##u = \frac{k+4i}{1+ki}## a purely real number. This falls under the subject area of complex numbers and their properties.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss calculating the complex conjugate of the expression and question whether solving for when the imaginary part equals zero is the correct approach. There is also a suggestion to test specific values of ##k## in the original expression.

Discussion Status

The discussion is active, with participants confirming their calculations and exploring the implications of their findings. There is an exchange of ideas regarding the method of using the conjugate and checking specific values.

Contextual Notes

Participants are working under the assumption that the expression must yield a purely real number, and there is a focus on the implications of the complex conjugate in this context.

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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.
 
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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.
Did you try setting ##k= \pm 2## in the original number to see whether you get a real number?

Did you mean you multiplied the denominator and numerator by the conjugate of the numerator?
 
Yes to the first and yes to the second (conjugate of the denominator though) and it looks all good!
 
squenshl said:
Yes to the first and yes to the second (conjugate of the denominator though) and it looks all good!
It does look good!
 

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