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

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To find values of k that make u = (k + 4i) / (1 + ki) a purely real number, the complex conjugate was calculated as (5k / (k^2 + 1)) + ((4 - k^2) / (k^2 + 1))i. Setting the imaginary part to zero, the equation (4 - k^2) / (k^2 + 1) = 0 leads to k = ±2. Substituting k = ±2 back into the original expression confirms that these values yield a real number. The calculations and approach to solving the problem were validated by participants in the discussion.
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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!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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