Given that $z = i + i^{2016} + i^{2017}$, find $|z^{10}|$.

Question

I am told that $$z = i + i ^ {2016} + i ^ {2017}$$

and I have to find $|z^{10}|$. This is what I tried:

$$i ^ {2016} = (i^4)^{504} = 1 ^ {504} = 1$$

$$i ^ {2017} = i \cdot (i^4)^{504} = i \cdot 1 = i$$

So we have that:

$$z = 1 + 2i$$

And then I worked towards finding $z ^ {10}$.

$$z = 1 + 2i$$

$$z^2 = 1 + 4i + 4i^2 = -3 + 4i$$

$$z^4 = ... = -7 - 24 i$$

$$z^5 = z^4 \cdot z = (-7-24i)(1 + 2i) = ... = 41 - 38i$$

$$z^{10} = (z^5)^2 = (41 - 38i)^2 = ... = 237 -3116i $$

So we then have:

$$|z^ {10} | = \sqrt{237^2 + 3116^2} = \sqrt{56169 + 964656} = \sqrt{1020825} = 15\sqrt{4537}$$

Is this correct? Is there a better way to solve this? More efficient/faster way?

Answer 1

$$|z^{10}|=|z|^{10}=(1^2+2^2)^5=3125.$$Your calculation contained an error: it should be $\sqrt{56159+9709456}$.

Answer 2

write in euler form for fast calculation $z=1+2i=\sqrt{5}\ e^{i\theta}$

$|z^{10}|=|z|^{10}=|\sqrt{5}\ e^{i\theta}|^{10}=5^5\times |e^{10i\theta}|=5^5$

Note : $\theta=\arctan2$

Answer 3

$z=1+2i$; $\overline{z}=1-2i$;

$|z|^2=z\cdot \overline{z}=1-4i^2=5$;

$|z|^{10}=(|z|^2)^5=5^5$.