Posted by jane on Friday, April 22, 2011 at 10:26pm.
Express all the solutions to the following equations in a + bi form:
(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 11:03pm
I'll do the first one:
z²+2z+4=0
Use the quadratic formula:
z=(2±√(2²16))/2
=1±sqrt(3)
=1±(√3)i
The remaining questions are similar.
For #c,
2z+iz=3i
z(2+i)=3i
z=(3i)/(2+i)
the expression must be rationalized by multiplying both numerator and denominator by the conjugate of the denominator, namely (2i).
I'll leave the details to you.
Post your answer for checking if you wish.

math  jane, Saturday, April 23, 2011 at 7:36am
thank you!
for part c I got 1i
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 1:08pm
Yes, for #2, you can use the results of #1, since the expressions are identical.
For #3, 1i is correct!
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