Exercises – Long Day#
Exercise 1: The complex exponential function#
Question a#
Use Euler’s formula (Equation (4.7) in the textbook) to write the following complex numbers in rectangular form, and draw the numbers in the complex plane:
\(\mathrm e^{i \frac{-\pi}{4}}\)
\(\mathrm e^{i\frac{\pi}{2}}\)
\(\mathrm e^{\pi i}\)
\(\mathrm e^{i \frac{5\pi}{4}}\)
What are the principal arguments of the numbers?
Answer
Its rectangular form is \(\frac12 \sqrt{2}- \frac12 \sqrt{2} i\) and its principal argument is \(-\pi/4\).
Its rectangular form is \(i\) and its principal argument is \(\pi/2\).
Its rectangular form is \(-1\) and its principal argument is \(\pi\).
Its rectangular form is \(-\frac12 \sqrt{2}- \frac12 \sqrt{2} i\). An argument can be read off directly as \(5\pi/4\), but its principal argument is \(-3\pi/4\).
This question illustrates the fact that for a real number \(t\), the complex number \(\mathrm e^{it}\) has modulus \(1\) and an argument \(t\).
Question b#
Use Definition 4.1.1 from the textbook to write the following complex numbers in rectangular form:
\(\mathrm e^{i\frac{\pi}{2}}\)
\(3\mathrm e^{1+\pi i}\)
Hint
Regarding the rectangular form of \(\mathrm e^{i\frac{\pi}{2}}\), since you are asked to use Definition 4.4.1, note that \(\mathrm e^{i\frac{\pi}{2}}=\mathrm e^{0+i\frac{\pi}{2}}\).
Answer
\(i\). This answer is of course the same as that for the second item in Question a. In fact, Euler’s formula (that is, Equation (4.7) in the textbook) is a special case of Definition 4.4.1 obtained by choosing \(a=0\) and \(b=t\).
\(-3\mathrm e\).
Exercise 2: Modulus and argument#
We are given the complex number \(w=1-i\).
Determine \(|\,w\,|\) and state an argument \(\arg(w)\).
Determine \(|\,\mathrm e^w\,|\) and state an argument \(\arg(\mathrm e^w)\).
Hint
Regarding an argument of \(\mathrm e^{1-i}\), a possible approach would be to use Definition 4.4.1 to write the number in rectangular form, and then to use Theorem 4.3.1 to determine the principal argument.
Hint
From the previous hint and Theorem 4.3.1, the principal argument of \(\mathrm e^{1-i}\) is found to be
Now remember that \(\tan(x)=\sin(x) / \cos(x)\) and that \(\mathrm{arctan}\) is the inverse function of \(\tan\).
Answer
\(|\,w\,|=\sqrt{2}\), and a possible argument is \(\arg(w)=-\frac{\pi}{4}\).
\(|\,\mathrm e^w\,|=\mathrm e\), and a possible argument is \(\arg(\mathrm e^w)=-1\).
In both cases, the chosen argument is in fact also the principal argument since both \(-\frac{\pi}{4}\) and \(-1\) lie in the interval \(]-\pi,\pi]\).
Exercise 3: Polar form#
This exercise builds on Exercise 5a from Short Day in Week 3. The numbers \(z_1=1+i\sqrt{3}\), \(z_2=-1+i\sqrt{3}\), \(z_3=-1-i\sqrt{3}\) and \(z_4=1-i\sqrt{3}\) are given.
Question a#
State the four numbers in polar form.
Hint
See Definition 4.6.1 for a reminder of what the polar form of a complex number is. You can reuse your results from Exercise 5a from Short Day in Week 3 to avoid repeating work.
Answer
\(z_1=2\mathrm e^{\frac{\pi}{3} i}\), \(z_2=2\mathrm e^{\frac{2\pi}{3} i}\), \(z_3=2\mathrm e^{\frac{-2\pi}{3} i}\), \(z_4=2\mathrm e^{\frac{-\pi}{3} i}\).
Question b#
Use the polar forms to compute \(z_1^{3}\), \(z_2^{3}\), \(z_3^{3}\) and \(z_4^{3}\).
Hint
You can use the last part of Theorem 4.6.2 in the textbook to calculate the modulus and an argument of an integer power of a complex number.
Answer
\(-8,8,8,-8.\)
Question c#
Show that \(z_2\) and \(z_3\) are roots of the polynomial \(Z^3-8\).
Hint
See Definition 5.1.2 to read more about what precisely a root of a polynomial is.
Question d#
Determine a polynomial \(p(Z)\) in \(\mathbb{C}[Z]\) of degree three that has \(z_1\) and \(z_4\) as roots.
Answer
\(Z^3+8\) is a valid answer, but there are other possibilities.
Exercise 4: First-degree polynomials#
A polynomial \(p(Z) \in \mathbb{C}[Z]\) is given by \(p(Z)=(2-i)Z+i\).
Question a#
Find the root of the polynomial \(p(Z)\).
Answer
\(\frac15-\frac25 i.\)
Question b#
Solve the polynomial equations \(p(z)=2\) and \(p(z)=-2+2i\).
Answer
The equation \(p(z)=2\) has the solution \(1\).
The equation \(p(z)=-2+2i\) has the solution \(-1\).
Exercise 5: Second-degree polynomials#
Question a#
Find all roots of the polynomial \(Z^2+2Z+5\).
Hint
Theorem 5.2.1 in the textbook explains how to find these roots.
Question b#
Find all roots of the polynomial \((Z^2+2Z+5)\cdot (Z^2-4)\).
Hint
You do not need to write the polynomial in expanded form by multiplying out the brackets. In fact, that would make it much harder for you to solve! Instead, consider what it means for a complex number to be a root of the polynomial \((Z^2+2Z+5)\cdot (Z^2-4)\). Remember the zero product rule from secondary school.
Answer
The polynomial has the four roots \(z_1=-1+2i\), \(z_2=-1-2i\), \(z_3=2\) and \(z_4=-2\).
Exercise 6: Polynomial arithmetic#
The following three polynomials in \(\mathbb{C}[Z]\) are given:
Question a#
Determine the degrees and leading coefficients of the three given polynomials.
Hint
Note that \(p_3(Z)\) is the same polynomial as \((1+i)Z^5-1\).
Answer
\(p_1(Z)\) has degree \(3\) and leading coefficient \(2\).
\(p_2(Z)\) has degree \(1\) and leading coefficient \(1\).
\(p_3(Z)\) has degree \(5\) and leading coefficient \(1+i\).
Question b#
Compute \(p_1(Z)+p_2(Z)+p_3(Z)\), \(ip_3(Z)\) and \(p_1(Z)p_2(Z)\).
Answer
\(p_1(Z)+p_2(Z)+p_3(Z)=(1+i)Z^5+2Z^3.\)
\(i \cdot p_3(Z)=(-1+i)Z^5-i.\)
\(p_1(Z)p_2(Z)=2Z^4+4Z^3-Z^2-3Z-2.\)
Exercise 7: Equations with the exponential function#
Question a#
We are given the numbers \(w_1=1\), \(w_2=\mathrm e\) and \(w_3=2i\). For \(n=1,2,3\), determine the solution set in \(\mathbb C\) to the equation
Hint
Lemma 4.6.1 in the textbook describes how solutions to an equation of the form \(\mathrm e^z=w\) are found.
Hint
Any argument of \(w\) equals the principal argument of \(w\) plus an integer multiple of \(2\pi\).
Answer
According to Lemma 4.6.1, each solution to the equation \(\mathrm e^z=1\) is of the form \(z=i \mathrm{arg}(1)\). The principal argument of \(1\) is \(0\), and all other arguments of \(1\) are found as this principal argument plus an integer multiple of \(2\pi\). Hence, the equation \(\mathrm e^z=1\) has the solution set \(\{ ip2\pi \, \mid \, p \in \mathbb{Z}\}.\)
\(\mathrm e^z=\mathrm e\) has the solution set \(\{ 1+ip2\pi \, \mid \, p \in \mathbb{Z}\}\).
\(\mathrm e^z=2i\) has the solution set \(\left\{ \ln(2)+i(\frac{\pi}{2}+p2\pi) \, \mid \, p \in \mathbb{Z}\right\}\).
Question b#
Determine the solution set of the equation
Answer
The solution set is the union of the solution sets to the equations in Question a that correspond to \(n=1\) and \(n=3\).
Question c#
Prove the first claim in Theorem 4.4.2, namely that \(\mathrm e^z \neq 0\) for all \(z\in\mathbb C\).
Hint
If you write \(z=a+bi\) in rectangular form, Definition 4.4.1 implies that \(\mathrm e^z=\mathrm e^a \cdot (\cos(b)+\sin(b) i)\). Can this expression ever be zero?
Exercise 8: Complex conjugation and roots of polynomials#
Question a#
Determine \(\overline{2-3i}\) and \(\overline{10+12i}\). State the answers in rectangular form.
Determine \(\overline{5 \mathrm e^{i\pi/3}}\). State the answer in polar form.
Hint
\(\overline{z}\) denotes the complex conjugate of a complex number \(z\); see Definition 4.2.3 in the textbook.
Hint
For item 2, Lemmas 5.3.1 and 5.3.2 in the textbook will be helpful.
Answer
\(\overline{2-3i}=2+3i\) and \(\overline{10+12i}=10-12i\).
\(\overline{5 \mathrm e^{i\pi/3}}=\overline{5} \overline{\mathrm e^{i\pi/3}}=5 \mathrm e^{-i\pi/3}\).
Question b#
We are informed that the complex number \(1+i\) is a root of the polynomial \(Z^3+(2+3i)Z+3-7i\). Show that \(1-i\) is a root of the polynomial \(Z^3+(2-3i)Z+3+7i\) using the properties of the complex conjugate as described in Lemma 5.3.1 in the textbook.
Hint
Since we know that \(1+i\) is a root of \(Z^3+(2+3i)Z+3-7i\), we also know that \((1+i)^3+(2+3i)(1+i)+3-7i=0\). What happens if you take the complex conjugate of the left- and right-hand sides?
Question c#
We are now being informed that the complex number \(1+i\) is a root of the polynomial \(Z^4+Z^2-2Z+6\). Show that \(1-i\) is also a root of this polynomial.
Hint
This time we know that \(1+i\) fulfills \((1+i)^4+(1+i)^2 -2(1+i)+6=0\). Again, investigate what happens if you take the complex conjugate of both side of the equation.
Exercise 9: Integer powers and polar form#
In Exercise 3 you wrote the complex number \(-1+\sqrt{3}i\) in polar form. Now, show that
Hint
Follow the method used in Example 4.6.2.
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.