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I was actually trying this simple question, might be just me being really rusty not doing maths for a very long time. I tried adding all the $\theta$ up on all zs but it doesn't seem to work.

Anyway as the title states,

$z_1,...z_n$ are the $n$ solutions of $z^n =a$ and $a$ is real number, show that $z_1+...+z_n$ is a real number.

Yau Kin Hoe
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    Are you allowed to use the fact that the roots of $z^n-a = 0$ are either real or occur in conjugate pairs? If so, the proof is trivial. If not, what are you permitted to use? – Brian Tung Nov 19 '15 at 08:09
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    In general, the sum of the solutions of p(z)=0 is equal to negative the coefficient of $x^{(n-1)}$, which in this case is 0. – jschnei Nov 19 '15 at 08:09

2 Answers2

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Two observations:

  1. If $z$ is a root of $x^n = a$, then $\overline{z}$ is as well, since $\overline{z}^n = \overline{(z^n)} = \overline{a} = a$.

  2. $z + \overline{z}$ is real.

Now do you see why it's true?

Alex Kruckman
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  • There is much simpler... – Did Nov 19 '15 at 08:10
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    @Did There are several very simple ways of seeing this. Whether one is much simpler seems to me a matter of opinion... – Alex Kruckman Nov 19 '15 at 08:16
  • yes i see why this is true but that does not answer the question. $z+\bar{z}$ is real does not mean $z_1+z_2$ is real – Yau Kin Hoe Nov 19 '15 at 08:34
  • Pair up each non-real root with its complex conjugate, which also appears on the list, and add them. Each time, you get a real number. Now you're left with a sum of real numbers. – Alex Kruckman Nov 19 '15 at 08:59
  • You see... Absolute relativism about matters of mathematical simplicity is not a very sustainable position. – Did Nov 19 '15 at 10:42
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Use thae fact that $$z^n -a =z^n -(z_1 +...+z_n ) z^{n-1} +...+ z_1 \cdot z_2 \cdot ...\cdot z_n $$