Find The Impedance For Two Complex Impedances in Parallel

[Ascendant0]

Homework Statement

Find the impedance of $z_1$ and $z_2$ in parallel: $z_1 = 2+3i$, $z_2 = 1-5i$

Relevant Equations

$(z_1^{-1} + z_2^{-1})^{-1}$

Finding the series for the first part of the problem was easy, but for parallel, I'm not sure how to separate the real from the imaginary in the fractions after I add them together?

So, I take: $(1/(2+3i) + 1/(1-5i)^{-1}$, and after I combine the denominators and combine all terms, I end up with:

$(17-7i)/(3-2i)$

From here, I'm not sure how to separate them? I see in the answer for this one they simply have $5+i$, but I don't see how to reduce what I have and separate the real and imaginary parts like that. Was there a different way I should've done this, or is there just some way to separate what I have above?

 

[Orodruin]

Homework Statement: Find the impedance of $z_1$ and $z_2$ in parallel: $z_1 = 2+3i$, $z_2 = 1-5i$
Relevant Equations: $(z_1^{-1} + z_2^{-1})^{-1}$

Finding the series for the first part of the problem was easy, but for parallel, I'm not sure how to separate the real from the imaginary in the fractions after I add them together?

So, I take: $(1/(2+3i) + 1/(1-5i)^{-1}$, and after I combine the denominators and combine all terms, I end up with:

$(17-7i)/(3-2i)$

From here, I'm not sure how to separate them? I see in the answer for this one they simply have $5+i$, but I don't see how to reduce what I have and separate the real and imaginary parts like that. Was there a different way I should've done this, or is there just some way to separate what I have above?

You can always make the denominator real by multiplying denominator and numerator by the conjugate of the denominator ($3+2i$).

 

[FactChecker]

I end up with:

$(17-7i)/(3-2i)$

Good work! Your answer is correct. All you need to do is to simplify it as @Orodruin suggests.

 

[Ascendant0]

You can always make the denominator real by multiplying denominator and numerator by the conjugate of the denominator ($3+2i$).

Thanks, yea I realized that this morning. I was half-asleep when I was doing this last night, and for some reason wasn't thinking of that. I appreciate the suggestion, and I did take care of that earlier, and got the $5 + i$ value.