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I have the following question. Let $P(z)$ be a complex polynomial. I'm looking to either prove or disprove that

$$\lim\limits_{\text{Re } z\to\infty} \text{Re } P(z) = \pm \infty.$$

It feels intuitively like it should be true, since $|P(z)|$ must go to infinity - except I can't find any way to prove or disprove it. I can't use $|P(z)|$ because the modulus relies on the imaginary part as well - which is where I got stuck.

If it's not true, though, then I wonder if it's possible to find a sequence such that the above limit holds?

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    It's easy if you write $z=x+iy$ and look at the dominant terms in the expansion. – 2017-02-06
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    Sorry, I don't get it. Would you mind explaining a bit more? – 2017-02-06

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If you take $P(z) = iz$, then clearly if you let $z$ go to infinity along the real axis then the limit of $Re(P(z))$ is zero. If you let it go to infinity along the line $Re(z) = Im(z)$, the limit is negative infinity. Note that in both cases $Re(z)$ is going to infinity, but $Re(P(z))$ approaches two different limits, so the limit does not exist and the conjecture is false.

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    It's not $z$ that's going to infinity, it's $Re(z)$, which is a real number, and therefore can not go to infinity along the line $Re(z)=Im(z)$. – 2017-02-06
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    @Marra: $Im(z)$ is also a real number. There is no problem with $Re(z)$ going to infinity along that diagonal line. The conjecture is that $|Re(P(z))|$ can be made arbitarily large by taking $Re(z)$ sufficiently large, and already considering $Re(z)\to 0$ while $Im(z) = 0$ proves this is not the case. – 2017-02-06
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    Yeah, I think you two are actually right, lemme think here – 2017-02-06
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    $Re(z)$ going to infinity is not the same thing as $z$ being real and going to infinity. – 2017-02-06
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    I didn't say that those are the same things. I just assumed that $Im(z)$ was meant to be zero, or at least constant. – 2017-02-06
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    @Marra: Even if it were assumed that $Im(z)$ were constant, the result would not be $\pm \infty$, it would be minus that constant. – 2017-02-06
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    Not in this example, in fact. – 2017-02-06
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    Hmm, thanks everyone! I think this works, and I think my intuition for complex analysis is really bad :P. This raises another interesting question (if one line gives a finite limit, is it always possible to find another which gives a different limit?), but I'll try that on my own first! – 2017-02-06