An algebra problem by Manish Mayank

Algebra Level 3

If p = √(-1 ) then find (1-p)^20

1024 -1024 512 -512

Manish Mayank by Manish Mayank , 20, India

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2 solutions

Zakaria Sellami
Jun 21, 2014

p = − 1 = i p=\sqrt{-1}=i then: ( 1 − p ) 20 = ( 1 − i ) 20 = ( 2 × e i π / 4 ) 20 = 2 20 e i 5 π = − 2 10 = − 1024 (1-p)^{20}=(1-i)^{20}=(\sqrt{2} \times e^{i \pi/4})^{20}=\sqrt{2}^{20}e^{i 5\pi}=-2^{10}=-1024

5 Helpful 0 Interesting 0 Brilliant 0 Confused
Hung Woei Neoh
Jul 5, 2016

p = − 1 = i ( 1 − p ) 20 = ( ( 1 − i ) 2 ) 10 = ( 1 − 2 i + i 2 ) 10 = ( 1 − 2 i − 1 ) 10 = ( − 2 i ) 10 = 1024 i 10 = 1024 ( i 4 ) ( i 4 ) ( i 2 ) = 1024 ( 1 ) ( 1 ) ( − 1 ) = − 1024 p=\sqrt{-1} = i\\ (1-p)^{20}\\ =\left((1-i)^2\right)^{10}\\ =\left(1-2i+i^2\right)^{10}\\ =\left(1-2i-1\right)^{10}\\ =(-2i)^{10}\\ =1024i^{10}\\ =1024(i^4)(i^4)(i^2)\\ =1024(1)(1)(-1)\\ =\boxed{-1024}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Same solution! :)

A Former Brilliant Member - 4 years, 11 months ago

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