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Find algerbracically the modulus and the argrument of the complex number...

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roshan-rox | (Level 1) Valedictorian

Posted June 4, 2013 at 4:35 PM via web

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Find algerbracically the modulus and the argrument of the complex number

`(-1+iota)^3/(1+iota)^4`

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted June 4, 2013 at 5:43 PM (Answer #1)

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`((-1+i)^3)/((1+i)^4)`   Expand the binomials:

`=(-1+3i-3i^2+i^3)/(1+4i+6i^2+4i^3+i^4)`

Simplify using `i=i,i^2=-1,i^3=-i,i^4=1` :

`=(2+2i)/(-4)`

`=-1/2-1/2i`

So `z=((-1+i)^3)/((1+i)^4)=-1/2-1/2i`

The modulus of `z=a+bi` is `|z|=sqrt(a^2+b^2)` so

`|z|=sqrt((-1/2)^2+(-1/2)^2)=sqrt(2)/2`

To find the argument we need the reference angle: `theta=tan^(-1)(b/a)` so `theta=tan^(-1)(1)=45^@`

Since a,b<0 the angle is in the third quadrant. Thus `arg(z)=225^@`

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`|z|=sqrt(2)/2,arg(z)=225^@`

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