Description
Problem 1
(2+3+5 points)
a) Find the (complex) roots of the polynomial p(x) = 2x
2 + 12x + 26.
b) Find the values of parameter b for which the equation bx2 − bx + 2 = 0 has no real roots.
c) Find all roots (real or complex) of the polynomial p(x) = x
6−x
5−3x
4−3x
3−22x
2+4x+24
.
Hint: x = 3 is a root. Divide out the associated linear factor and continue with more
roots that are easy to guess.
Problem 2
(3+3+4 points and (2+3) bonus points)
Assuming that z = a + ib is a complex number, compute real and imaginary parts of
a) 1
(z
∗)
2
b) 2+z
2z+2
c) (z
∗
)
2
z
Bonus: |x| is the absolute value function:
In the case of x ∈ C: |x| =
√
xx∗
In the case of x ∈ R: |x| =
√
x
2 or in other words |x| = x if x ≥ 0; |x| = −x if x < 0.
In both cases |x| ∈ R and |x| ≥ 0.
d) Compute |
1−i
2+i
| . Use the definition of the absolute value function for complex numbers.
d) Characterize the set of real numbers x that satisfy |4x + 2| ≤ |2x − 3| .
Hint: You cannot directly work with | |. Use the definition of absolute value for real
numbers to change the inequality into an equivalent problem without | |. For that, you
can apply certain functions to both sides of the inequality without changing the inequality.
Problem 3
(4+3+3 points)
Proof the following for complex numbers z and w, i.e. z, w ∈ C.
a) z
∗
w∗ = ( z
w
)
∗
b) Re(z) = z+z
∗
2
c) Im(z) = z−z
∗
2i
Re(z) and Im(z) are the real and complex part of z, respectively. I.e. if z = a + bi, then
Re(z) = a and Im(z) = b.
