Given $z = {({(i)}^{i})}^{i}$ $z=((i)^i)^i$ calculate $| z |$ $abs z$ ?

Question

Given $z = {({(i)}^{i})}^{i}$ $z=((i)^i)^i$ calculate $| z |$ $abs z$ ?

2 Answers

Impact of this question

1635 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-29T04:41:13

$| z | = 1$ $|z|=1$

Explanation:

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

$cos (\frac{π}{2}) + i sin (\frac{π}{2}) = e^{i \frac{π}{2}}$ $cos(pi/2)+isin(pi/2)=e^(ipi/2)$

Hence $z = {({(ι)}^{ι})}^{ι}$ $z=((iota)^iota)^iota$

= ${({(e^{i \frac{π}{2}})}^{i})}^{i}$ $((e^(ipi/2))^i)^i$

= ${(e^{\frac{π}{2} i^{2}})}^{i}$ $(e^(pi/2i^2))^i$

= ${(e^{- \frac{π}{2}})}^{i}$ $(e^(-pi/2))^i$

= $e^{- \frac{π}{2} i}$ $e^(-pi/2i)$

= $cos (- \frac{π}{2}) + i sin (- \frac{π}{2})$ $cos(-pi/2)+isin(-pi/2)$

= $0 - i$ $0-i$

= $- i$ $-i$

and $| z | = 1$ $|z|=1$

Answer 2 · 2017-04-29T12:51:09

$1$ $1$

Explanation:

$| z | = \sqrt{¯ z \cdot z}$ $absz =sqrt( bar z cdot z)$

now

$z = {({(i)}^{i})}^{i}$ $z=((i)^i)^i$ and

$¯ z = {({(- i)}^{- i})}^{- i}$ $bar z = ((-i)^-i)^-i$

then

$¯ z \cdot z = {({(- i)}^{- i})}^{- i} {({(i)}^{i})}^{i} = {(- i)}^{- i} {(i)}^{i} = (- i) i = 1$ $bar z cdot z = ((-i)^-i)^-i((i)^i)^i = (-i)^-i(i)^i=(-i)i=1$

and

$| z | = \sqrt{1} = 1$ $absz = sqrt1 = 1$

Science

Math

Humanities

... and beyond

Given $z = {({(i)}^{i})}^{i}$ $z=((i)^i)^i$ calculate $| z |$ $abs z$ ?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

Given z=((i)i)iz=((i)^i)^i calculate |z|abs z ?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Given $z = {({(i)}^{i})}^{i}$ $z=((i)^i)^i$ calculate $| z |$ $abs z$ ?