Hello, I am not sure this question is meant to go here. This is a very simple problem, and I know how to solve it the easier way, I am struggling to find my error when solving it in a harder way.
Here is the problem
Easy way to solve is just to multiply both sides by z and then find the conjugate for denominator and solved. The answer is
However, I am thinking why this does not work:
Let \(\displaystyle \displaystyle{z=\mu+\alpha i}\), then we have
Now expand:
This means that
We have two equations and two unknowns, thus we can solve it by substitution or by inspection.
Now I believe that the following step is incorrect, and I need to understand why:
You then have a system of equations:
By inspection \(\displaystyle \mu = \frac{5}{2}\) and thus \(\displaystyle \alpha = \frac{45}{16}\)
But clearly this is incorrect because when you substitute these into fraction with the \(\displaystyle i\), you are not getting \(\displaystyle -i\). Therefore that step must be incorrect. If so, why is it incorrect? Or is something else incorrect?
Thank you
Here is the problem
\(\displaystyle \displaystyle{\frac{5+2i}{z}=2-i}\)
Easy way to solve is just to multiply both sides by z and then find the conjugate for denominator and solved. The answer is
\(\displaystyle \displaystyle{\frac{8+9i}{5}}\)
However, I am thinking why this does not work:
Let \(\displaystyle \displaystyle{z=\mu+\alpha i}\), then we have
\(\displaystyle \displaystyle{\frac{5+2i}{\mu + \alpha i} * \frac{\mu - \alpha i}{\mu - \alpha i}}\)
Now expand:
\(\displaystyle \displaystyle{\frac{5\mu + i(2\mu - 5\alpha) + 2\alpha}{\mu^2 + \alpha}}\), this is meant to equal to \(\displaystyle 2-i\)
This means that
\(\displaystyle \displaystyle{\frac{5\mu + 2\alpha}{\mu^2 + \alpha} = 2}\) and \(\displaystyle \displaystyle{\frac{i(2\mu - 5\alpha)}{\mu^2+\alpha}=-i}\)
We have two equations and two unknowns, thus we can solve it by substitution or by inspection.
Now I believe that the following step is incorrect, and I need to understand why:
\(\displaystyle \displaystyle{\frac{i(2\mu - 5\alpha)}{\mu^2+\alpha}=-i \Rightarrow -i(\mu^2+\alpha)=i(2\mu - 5\alpha) \Rightarrow -\mu^2-\alpha=2\mu - 5\alpha}\)
You then have a system of equations:
\(\displaystyle \begin{eqnarray} -\mu^2-\alpha = 2\mu - 5\alpha\\ 2\mu^2+2\alpha = 5\mu + 2\alpha \end{eqnarray} \)
By inspection \(\displaystyle \mu = \frac{5}{2}\) and thus \(\displaystyle \alpha = \frac{45}{16}\)
But clearly this is incorrect because when you substitute these into fraction with the \(\displaystyle i\), you are not getting \(\displaystyle -i\). Therefore that step must be incorrect. If so, why is it incorrect? Or is something else incorrect?
Thank you
