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I'm currently working on a problem and I need to simplify the following complex quantity in the form of $a+bi$. Assuming that $a$ and $b$ are real numbers, here is the quantity:

$${(a+bi)\over (a-bi)}- {(a-bi)\over (a+bi)}$$

I've rationalized the left and the right terms and arrived at an answer of $\dfrac{4abi}{a^2 + b^2}$, but do not know where to go from here. I double checked my math and everything should be correct up to this point. What's the problem here?

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    Also, please tell me if the question is hard to read. The font may be a bit small. – 2017-02-02
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    Why do you think that you should be able to simplify further? – 2017-02-02
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    Well I don't see how else I can reduce it. – 2017-02-02

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You have $z=a+bi$ and want to compute $$\frac{z}{\bar z}-\frac{\bar z } {z}$$ Where $\bar z= a-bi$. Combine denominators to get $$\frac{z^2-\bar z^2}{z\bar z}$$ From which you get $$\frac{4iab}{a^2+b^2}$$ So if you wanted to put this in the shape of a complex number $A+Bi$ you get that $A=0$ and $B=\frac{4ab}{a^2+b^2}$.

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    Okay I understand that but can you briefly explain why A = 0? – 2017-02-02
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    @Oliver821 complex numbers can be expressed, as you said, as$A+Bi$ where $A,B\in\mathbb{R}$ and $i^2=-1$. $0$ is a real number and so is $\frac{4ab}{a^2+b^2}$ since both $a$ and $b$ are real numbers (multiplication of real numbers yields a real and division of two real numbers where the denominator is not zero yields another real). So both $A=0$ and $B=\frac{4ab}{a^2+b^2}$ satisfy the conditions. – 2017-02-02
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    Now I see it! Thanks for the explanation. – 2017-02-02