Show that $z^3 + (1+i)z - 3 + i = 0$ does not have any roots in the unit circle $|z|\leq 1$.

I need help with showing that $z^3 + (1+i)z - 3 + i = 0$ does not have any roots in the unit circle $|z|\leq 1$?

My approach so far has been to try to develop the expression further.

$$z^3 +(1+i)z-3+i = z(z^2+i+1)-3+i$$

$$z(z^2+i+1) = 3 - i \longrightarrow |z(z^2+i+1)| = |3 - i|$$

This gives me the expression:

$z((z^2+1)^2+(1)^2) = \sqrt{10}$

Which can be written as:

$z(z^4 +2z^2 +2) = \sqrt{10}$

But how do I move on from here? Or am I attempting the wrong solution?

Thank you for your help!

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You used that $z\in\Bbb R^+$ though $z\in\Bbb C$, it would be better to write your expression as $$|z|^2\cdot|z+1+i|^2=10\ .$$ –  Berci Mar 5 '13 at 21:50

$z^3 + (1+i)z - 3 + i = 0\iff z^3+(1+i)z=3-i$

Now, If $|z|\leq 1$, then $|z^3+(1+i)z|\leq |z|^3+|1+i||z|\leq1+\sqrt2$

As $|3-i|=\sqrt{10}\gt 1+\sqrt{2}$

Therefore, $z^3+(1+i)z\neq 3-i$ for any $z\in \Bbb C, |z|\leq 1$

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Thank you for the excellent answer Avatar! –  Lukas Arvidsson Mar 6 '13 at 8:00
:):):):):):):):) –  Aang Mar 6 '13 at 8:01
If you write it as $z^3+(1+i)z=3-i$ (as you did), note that for $|z| \le 1$, the first term has modulus no greater than $1$ and the second no greater than $\sqrt 2$ and the triangle inequality solves your problem.
You can think of $z^3, (1+I)z,$ and $i-3$ as three vectors in $\mathbb R^2$ that add to $0$. So $|z^3|+|z(1+i)|\ge|i-3|$ and you know $|z|\le 1$ –  Ross Millikan Mar 5 '13 at 22:16