Real Value of x in Complex Equation?

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To find the real value of x in the equation Re(10 / (x + 4i)) = 1, the equation can be transformed into (10x - 40i) / (x^2 + 16) = 1. The real part of this expression is 10x / (x^2 + 16), while the imaginary part is -40 / (x^2 + 16). To solve for x, set the real part equal to 1, leading to the equation 10x / (x^2 + 16) = 1. This allows for solving x by isolating it in the equation.
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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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\frac{10x-40i}{x^2+ 16}= \frac{10x}{x^2+16}-i\frac{40}{x^2+ 16}
 
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.
 
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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