I'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.
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!
3 answers
thank 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?
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?
Yes, for #2, you can use the results of #1, since the expressions are identical.
For #3, 1-i is correct!
For #3, 1-i is correct!