Exercises – Short Day#
Exercise 1: Polynomials with real coefficients#
Question a#
Check, without using the quadratic formula, that \(-1+2i\) is a root of the polynomial \(3Z^2+6Z+15\).
Hint
Definition 5.1.2 explains how to check whether a given complex number is a root.
Question b#
Find another root of the polynomial \(3Z^2+6Z+15\), still without using the quadratic formula.
Hint
The last part of Section 5.3 might be useful.
Answer
Since the polynomial is real (has real coefficients), the complex conjugate of the root \(-1+2i\) is also a root of the polynomial. In other words, \(-1-2i\) is also a root of \(3Z^2+6Z+15\).
Exercise 2: Binomial equations#
Solve the binomial equation \(z^3=-8i\). Provide the solutions in rectangular form.
Hint
Binomial equations are equations of the form \(z^n=w\). These are treated in Theorem 5.4.1. In our case, we have \(n=3\) and \(w=-8i\).
Answer
\(z^3=-8i\) has the solutions \(2i\), \(\sqrt{3}-i\) and \(-\sqrt{3}-i\).
Exercise 3: A second-degree polynomial#
Find all complex roots of the polynomial \(Z^2+i\). State the roots in both polar and rectangular form.
Hint
A root \(z\) of the polynomial \(Z^2+i\) satisfies the equation \(z^2+i=0\). This equation can be rewritten as a binomial equation.
Hint
The polar form of the number \(-i\) is \(1 \cdot \mathrm e^{-i \pi/2}\).
Answer
The roots of \(Z^2+i\) in polar form are \(\mathrm e^{-i\pi/4}\) and \(\mathrm e^{i3\pi/4}\). In rectangular form they are \(\sqrt{2}/2-i\sqrt{2}/2\) and \(-\sqrt{2}/2+i\sqrt{2}/2\).
Exercise 4: Binomial second-degree equations with real right-hand sides#
Question a#
Let \(r\) be a positive real number. Use Theorem 5.2.1 to show that the equation
has exactly two solutions, which are given by \(-i\sqrt{r}\) and \(i\sqrt{r}\).
Question b#
Solve Question a again, this time using Theorem 5.4.1.
Hint
When using Theorem 5.4.1, which treats the general form \(z^n=w\), we have \(n=2\) and \(w=-r\).
Hint
What is the principal argument of a negative real number?
Question c#
Solve the equation \(z^2=-16\).
Answer
\(z^2=-16\) has the solutions \(-4i\) and \(4i\).
Exercise 5: The exponential function in a binomial equation#
In this exercise we wish to solve the equation \((\mathrm e^z)^4=1\) for \(z\) in the complex numbers using two different methods.
Question a – Method 1#
Let \(w=\mathrm e^z\) so that you can rewrite the equation \((\mathrm e^z)^4=1\) to \(w^4=1\). Now solve, first for \(w\) and then for \(z\).
Hint
The equation \(w^4=1\) is a binomial equation. Its right-hand side can be written as \(1=1\cdot \mathrm e^{0i}\).
Hint
The binomial equation \(w^4=1\) has the solutions \(w_1=1\), \(w_2=i\), \(w_3=-1\) and \(w_4=-i\). Hence we have that \((\mathrm e^z)^4=1\) if and only if \(\mathrm e^z=1\) or \(\mathrm e^z=i\) or \(\mathrm e^z=-1\) or \(\mathrm e^z=-i\). These four equations can be solved using Lemma 4.6.1.
Answer
\(\mathrm e^z=1\) has the solutions \(z=2\pi p i\), where \(p \in \mathbb{Z}\).
\(\mathrm e^z=i\) has the solutions \(z=(\pi/2+2\pi p)i\), where \(p \in \mathbb{Z}\).
\(\mathrm e^z=-1\) has the solutions \(z=(\pi+2\pi p)i\), where \(p \in \mathbb{Z}\).
\(\mathrm e^z=-i\) has the solutions \(z=(-\pi/2+2\pi p)i\), where \(p \in \mathbb{Z}\).
The solutions are therefore all complex numbers \(z\) that can be written in one of these four ways.
Question b – Method 2#
Use Theorem 4.4.2 to rewrite the equation \((\mathrm e^z)^4=1\) as \(\mathrm e^{4z}=1\). Then use Lemma 4.6.1 to find all solutions, and check that these match those you found in Question a.
Answer
The solutions are \(z=(2\pi p i)/4=\pi p/2 i\), where \(p \in \mathbb{Z}\). As \(p\) runs through all integers, we obtain the same solutions as in Question a.
Have you finished all the exercises?#
If it went too quickly and you are eager for more, then click here for some challenging optional extra exercises.