Given that $a+bi=p$ and $a^2-b^2=q$, how can I compute $ab+i$?
Note
$i=\sqrt{-1}$ is the complex number
$a$ and $b$ can be any integer
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$\begingroup$ In terms of $a$ and $b$, $ab+i$ is $ab+i$... $\endgroup$ – Arnaud Mortier Mar 2 at 13:48
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$\begingroup$ Hint: what is $\frac{p^2-q}{2i}$? $\endgroup$ – Conrad Mar 2 at 13:50
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$\begingroup$ I assume that you mean to compute it in terms of $p$ and $q$ not in terms of $a$ and $b$. $\endgroup$ – quid♦ Mar 2 at 16:28
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$$p^2=a^2+2iab-b^2=q+2iab$$ Therefore $$ab+i=\frac{p^2-q}{2i} +i$$
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Squaring the firtst equation we get $$a^2-b^2+2abi=p^2$$ so we get $$ab=-\left(\frac{p^2-q^2}{2}\right)i$$ Can you finish? We get $$ab+i=\left(-\left(\frac{p^2-q^2}{2}\right)+1\right)i$$
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Square both sides, so $(a)^2+(bi)^2+2abi=p^2$. Since $i^2=-1$, $a^2-b^2+2abi=p^2$.
Substituting $q=a^2-b^2$, $q+2abi=p^2$, and $ab=\frac{p^2-q}{2i}$, so $ab+i=\frac{p^2-1}{2i}+i=\frac{p^2-q}{2i}+\frac{2i^2}{2i}=\frac{p^2-q-2}{2i}$
