# Given z=((i)^i)^i calculate abs z ?

Apr 29, 2017

$| z | = 1$

#### Explanation:

$\iota = \sqrt{- 1}$ can also be written as $0 + i$ or

$\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right) = {e}^{i \frac{\pi}{2}}$

Hence $z = {\left({\left(\iota\right)}^{\iota}\right)}^{\iota}$

= ${\left({\left({e}^{i \frac{\pi}{2}}\right)}^{i}\right)}^{i}$

= ${\left({e}^{\frac{\pi}{2} {i}^{2}}\right)}^{i}$

= ${\left({e}^{- \frac{\pi}{2}}\right)}^{i}$

= ${e}^{- \frac{\pi}{2} i}$

= $\cos \left(- \frac{\pi}{2}\right) + i \sin \left(- \frac{\pi}{2}\right)$

= $0 - i$

= $- i$

and $| z | = 1$

Apr 29, 2017

$1$

#### Explanation:

$\left\mid z \right\mid = \sqrt{\overline{z} \cdot z}$

now

$z = {\left({\left(i\right)}^{i}\right)}^{i}$ and

$\overline{z} = {\left({\left(- i\right)}^{-} i\right)}^{-} i$

then

$\overline{z} \cdot z = {\left({\left(- i\right)}^{-} i\right)}^{-} i {\left({\left(i\right)}^{i}\right)}^{i} = {\left(- i\right)}^{-} i {\left(i\right)}^{i} = \left(- i\right) i = 1$

and

$\left\mid z \right\mid = \sqrt{1} = 1$