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

2 Answers
Shwetank Mauria
Apr 29, 2017

|z|=1

Explanation:

iota=sqrt(-1) can also be written as 0+i or

cos(pi/2)+isin(pi/2)=e^(ipi/2)

Hence z=((iota)^iota)^iota

= ((e^(ipi/2))^i)^i

= (e^(pi/2i^2))^i

= (e^(-pi/2))^i

= e^(-pi/2i)

= cos(-pi/2)+isin(-pi/2)

= 0-i

= -i

and |z|=1

Cesareo R.
Apr 29, 2017

1

Explanation:

absz =sqrt( bar z cdot z)

now

z=((i)^i)^i and

bar z = ((-i)^-i)^-i

then

bar z cdot z = ((-i)^-i)^-i((i)^i)^i = (-i)^-i(i)^i=(-i)i=1

and

absz = sqrt1 = 1

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