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December 18, 2014

December 18, 2014

Posted by **jane** on Friday, April 22, 2011 at 10:26pm.

(a) z^2 +2z+4 = 0

(b) z^3 −8 = 0 [Hint: z^3 −8 = (z−2)(z^2 +2z+4)]

(c) 2z+iz = 3−i

thank you!

- math -
**MathMate**, Friday, April 22, 2011 at 11:03pmI'll do the first one:

z²+2z+4=0

Use the quadratic formula:

z=(-2±√(2²-16))/2

=-1±sqrt(-3)

=-1±(√3)i

The remaining questions are similar.

For #c,

2z+iz=3-i

z(2+i)=3-i

z=(3-i)/(2+i)

the expression must be rationalized by multiplying both numerator and denominator by the conjugate of the denominator, namely (2-i).

I'll leave the details to you.

Post your answer for checking if you wish.

- math -
**jane**, Saturday, April 23, 2011 at 7:36amthank you!

for part c I got 1-i

but for part b, am I supposed to use the answer i got in part a to answer the question?

- math -
**MathMate**, Saturday, April 23, 2011 at 1:08pmYes, for #2, you can use the results of #1, since the expressions are identical.

For #3, 1-i is correct!

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