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If $a$ and $b$ are real and transcendental (over $\mathbb Q$), does it follow that $a+bi$ is also transcendental? I tried looking for a counterexample, but I don't actually know of many transcendental numbers besides $e$ and $\pi$, and I can't tell if, say, $e+i\pi$ is algebraic. Thus, I'm assuming that the statement is true, but is it? And how to prove it?

Also, what if $a$ is transcendental and $b$ is algebraic. Must $a+bi$ be transcendental?

Nishant
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    If $z=a+bi$ were algebraic, then so would $\overline{z}=a-bi$. Then as a sum/difference of two algebraics, the numbers $$a=\frac{z+\overline{z}}2,\qquad b=\frac{z-\overline{z}}{2i}$$ would also be algebraic. – Jyrki Lahtonen May 10 '14 at 18:50
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    For the second question, if $z=a+bi$ is algebraic and $b$ is algebraic, then $a=z-i\cdot b$ is also algebraic. – Hagen von Eitzen May 10 '14 at 18:53

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This question has been answered by Jyrki Lahtonen and Hagen von Eitzen in the comments.

If $z=a+bi$ were algebraic, then so would $\bar z=a-bi$. Then as a sum/difference of two algebraics, the numbers $$a=\frac{z+\bar z}2, \qquad b=\frac{z-\bar z}{2i}$$ would also be algebraic.

– Jyrki Lahtonen

For the second question, if $z=a+bi$ is algebraic and $b$ is algebraic, then $a=z-i\cdot b$ is also algebraic.

– Hagen von Eitzen

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Jim Belk
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