# Roots of complex numbers

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### #1HeatherL

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Posted 19 May 2009 - 08:52 AM

Hi!!
Could somebody pleeeeeease help me with this question....
Solve the equation z = 8
Normally i'm fine with this type of question but this one just stumped me and i'm starting to panic
Thanks,
Heather

### #2Marcus

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Posted 19 May 2009 - 02:17 PM

rearrange into z³-8=0, now from this you can see the z=2 must be a solution, (as sticking z=2 makes the equation true)

so synthetic division (or long division if you want to waste time )
CODE
|1   0   0  -8
2  |    2   4   8
|_______________
1   2   4   0

so z³-8=(z-2)(z²+2z+4)
so we need the roots of the quadratic , back to good 'al quadratic formula

a=1,b=2,c=4

Working
so the solutions are z=2, 1+i 3 , 1-i 3
=-=-=Marcus=-=-=

### #3HeatherL

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Posted 20 May 2009 - 09:36 AM

ooooo I get it now !!!thanks!!!!!!!!
Heather

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