Are irrational complex numbers possible?
Andrew Mclaughlin 
I am asking because I was reading this and the mathematics is a little over my head. The title of the paper is Rational Approximations to Irrational Complex Number, and I didn't think that complex irrational numbers could exist.
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$\begingroup$The paper defines rational complex numbers as numbers of the form $$x=\frac{a+bi}{c+di} \text{ where } a,b,c,d\in\mathbb Z.$$ It is easy to see that irrational complex numbers (complex numbers not of the above form) exist. For one thing, any number of the above form has norm of the form $$\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}$$ which is either rational or quadratic irrational yet (for example) $|\sqrt[3]{2}+i|=\sqrt{\sqrt[3]{2}+1}$ which is of algebraic degree $6$.
Edit: Corrected statment about norms. Thank you Michael Boratko for pointing out my error.
$\endgroup$ 24 $\begingroup$Taking the definition of a complex rational number $z$ as any number which can be represented by the form:
$$z= \frac {a+bi}{c+di}\qquad \text{for } a,b,c,d\in\mathbb Z$$
it is easy to see that if $z$ is any number with irrational parts (say, for instance, $\sqrt 2 + i$ as Alex mentions, or even just an irrational number like $\pi$) then it is also an irrational complex number. This is because we have $$\frac {a+bi}{c+di}=\frac {(a+bi)(c-di)}{(c+di)(c-di)}=\frac{ac+bd}{c^2+d^2}+\frac{cd-da}{c^2+d^2}i\, ,$$ and thus when we equate real and imaginary parts...
Indeed, one can verify that all complex rational numbers are just the points in the complex plane with purely rational coordinates. The definition may just as well have been all numbers $z$ which can be written in the form $$z=\frac a b + \frac c d i \qquad \text {for } a, b, c, d \in \mathbb Z$$ or just $$z=p+qi\qquad \text {for } p,q \in \mathbb Q$$
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