# Problem 1

- Find the (complex) roots of the polynomial
*p*(*x*) = 2*x*^{2 }+ 12*x*+ 26. - Find the values of parameter
*b*for which the equation*bx*^{2 }−*bx*+ 2 = 0 has no real roots. - Find all roots (real or complex) of the polynomial
*p*(*x*) =*x*^{6}−*x*^{5}−3*x*^{4}−3*x*^{3}−22*x*^{2}+4*x*+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

Assuming that *z *= *a *+ *ib *is a complex number, compute real and imaginary parts of a)

b)

- (
*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.

- Compute . Use the definition of the absolute value function for complex numbers.
- d) Characterize the set of real numbers
*x*that satisfy |4*x*+ 2| ≤ |2*x*− 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

Proof the following for complex numbers *z *and *w*, i.e. *z,w *∈ C. a)

- Re(
- Im(

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*.

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