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!
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.
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?
Yes, for #2, you can use the results of #1, since the expressions are identical.
For #3, 1-i is correct!
(a) To express the solutions of the equation z^2 + 2z + 4 = 0 in the form a + bi, we can use the quadratic formula.
The quadratic formula states that for any quadratic equation of the form ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac))/(2a)
In this case, the equation is z^2 + 2z + 4 = 0, so we have a = 1, b = 2, and c = 4.
Substituting these values into the quadratic formula, we get:
z = (-2 ± √(2^2 - 4(1)(4)))/(2(1))
Simplifying further, we have:
z = (-2 ± √(4 - 16))/2
z = (-2 ± √(-12))/2
Since we have a negative value inside the square root, it indicates that the solutions will be complex numbers. To express them in the form a + bi, we can write:
z = (-2 ± 2i√(3))/2
Simplifying further, we have:
z = -1 ± i√(3)
So the solutions to the equation z^2 + 2z + 4 = 0 in the form a + bi are:
z1 = -1 + i√(3)
z2 = -1 - i√(3)
(b) Given the equation z^3 - 8 = 0, we can factorize it using the difference of cubes formula: a^3 - b^3 = (a - b)(a^2 + ab + b^2).
In this case, we have a = z and b = 2, so we can rewrite the equation as:
z^3 - 8 = (z - 2)(z^2 + 2z + 4) = 0
From here, we have two possible solutions:
z - 2 = 0
This gives us z = 2.
z^2 + 2z + 4 = 0
To solve this quadratic equation, we can use the quadratic formula, similar to part (a).
Substituting a = 1, b = 2, and c = 4 into the quadratic formula, we get:
z = (-2 ± √(2^2 - 4(1)(4)))/(2(1))
Simplifying further, we have:
z = (-2 ± √(-12))/2
Since we have a negative value inside the square root, the solutions will be complex numbers. Expressing them in the form a + bi, we have:
z = -1 ± i√(3)
So the complete solutions to the equation z^3 - 8 = 0 in the form a + bi are:
z1 = 2
z2 = -1 + i√(3)
z3 = -1 - i√(3)
(c) To solve the equation 2z + iz = 3 - i, we can rearrange the terms to isolate z on one side.
First, let's simplify the equation:
2z + iz = 3 - i
Combine like terms:
z(2 + i) = 3 - i
Now, divide both sides by (2 + i):
z = (3 - i)/(2 + i)
To rationalize the denominator, we multiply the numerator and denominator by the conjugate of (2 + i), which is (2 - i):
z = [(3 - i)(2 - i)]/[(2 + i)(2 - i)]
Expanding the numerator and denominator:
z = (6 - 2i - 3i + i^2)/(4 - 2i + 2i - i^2)
Simplifying further:
z = (6 - 5i + i^2)/(4 + 1)
Since i^2 is defined as -1, we can substitute:
z = (6 - 5i - 1)/(4 + 1)
z = (5 - 5i)/5
Simplifying:
z = 1 - i
So the solution to the equation 2z + iz = 3 - i in the form a + bi is:
z = 1 - i