Exercises – Long Day

Exercises – Long Day#

Exercise 1: The Complex Exponential Function#

Question a#

Rewrite the following complex numbers to rectangular form by using Euler’s formula (Equation (4.7) in the textbook), and draw the numbers in the compex plane:

  1. \(\mathrm e^{i \frac{-\pi}{4}}\)

  2. \(\mathrm e^{i\frac{\pi}{2}}\)

  3. \(\mathrm e^{\pi i}\)

  4. \(\mathrm e^{i \frac{5\pi}{4}}\)

What is the principal argument of the numbers?

answer

  1. Its rectangular form is \(\frac12 \sqrt{2}- \frac12 \sqrt{2} i\) and its principal argument is \(-\pi/4\).

  2. Its rectangular form is \(i\) and its principal argument is \(\pi/2\).

  3. Its rectangular form is \(-1\) and its principal argument is \(\pi\).

  4. Its rectangular form is \(-\frac12 \sqrt{2}- \frac12 \sqrt{2} i\). An argument is read directly to be \(5\pi/4\), but its principal argument is \(-3\pi/4\).

This question illustrates that for a real number \(t\), the complex number \(\mathrm e^{it}\) has a modulus \(1\) and an argument of \(t\).

Question b#

Rewrite the following complex numbers to rectangular form by using Definition 4.4.1 from the textbook:

  1. \(\mathrm e^{i\frac{\pi}{2}}\)

  2. \(3\mathrm e^{1+\pi i}\)

Hint

Regarding the rectangular form of \(\mathrm e^{i\frac{\pi}{2}}\), now that you are being asked to use Definition 4.4.1 then note that \(\mathrm e^{i\frac{\pi}{2}}=\mathrm e^{0+i\frac{\pi}{2}}.\)

answer

  1. \(i\). This answer is of course the same as that for the second bullet 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 achieved by choosing \(a=0\) and \(b=t\).

  2. \(-3\mathrm e\).

Exercise 2: Modulus and Argument#

We are given the complex number \(w=1-i\,\).

  1. Determine \(|\,w\,|\) and state an argument \(\arg(w)\,\).

  2. 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 rewrite the number to its 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

\[\mathrm{arctan}\left(\frac{\mathrm e\sin(-1)}{\mathrm e\cos(-1)}\right).\]

Now remember that \(\tan(x)=\sin(x) / \cos(x)\) and that \(\mathrm{arctan}\) is the inverse function of \(\tan\).

Answer

  1. \(|\,w\,|=\sqrt{2}\), and a possible argument is \(\arg(w)=-\frac{\pi}{4}\,.\)

  2. \(|\,\mathrm e^w\,|=\mathrm e\), 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 \(-1\) are found within the interval \(]-\pi,\pi]\).

Exercise 3: Polar Form#

This exercise builds further upon Exercise 5a from the Short Day of 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 to brush up on what the polar form of a complex number is all about. You can reuse your results from Exercise 5a from Short Day of Week 3 to avoid double 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 from the textbook to compute modulus and 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, and there are more options.

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 a 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 from the textbook tells how these roots are found.

Question b#

Find all roots of the polynomial \((Z^2+2Z+5)\cdot (Z^2-4)\).

Hint

You do not have to multiply the brackets together and thereby expand the polynomial. In fact, that would make it much harder for you to solve! Rather, 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 rule of zero product from highschool.

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 Arithmetics#

The following three polynomials in \(\mathbb{C}[Z]\) are given:

\[p_1(Z)=2Z^3-Z-1,\]
\[p_2(Z)=2+Z,\]
\[p_3(Z)=-1+0Z^{10}+(1+i)Z^5.\]

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 \(n=1,\dots,3\), determine the solution set (the set containing all solutions) in \(\mathbb C\) of the equations:

\[\mathrm e^z=w_n.\]

Hint

Lemma 4.6.1 in the textbook describes how solutions to an equation of the form \(\mathrm e^z=w\) are found.

Hint

An arbitrary 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

\[(\mathrm e^z-1)(\mathrm e^z-2i)=0\,.\]

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, that being 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#

  1. Determine \(\overline{2-3i}\) and \(\overline{10+12i}\). State the answers in rectangular form.

  2. 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 bullet 2, Lemmas 5.3.1 and 5.3.2 in the textbook will be of help.

Answer

  1. \(\overline{2-3i}=2+3i\) and \(\overline{10+12i}=10-12i\).

  2. \(\overline{5 \mathrm e^{i\pi/3}}=\overline{5} \overline{\mathrm e^{i\pi/3}}=5 \mathrm e^{-i\pi/3}\).

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 \(Z^3+(2-3i)Z+3+7i\) by utilising 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 \(1-i\) also is a root of this polynomial.

Hint

This time we are being given that \(1+i\) fulfills \((1+i)^4+(1+i)^2 -2(1+i)+6=0.\) Again, investigate what happens if the complex conjugate is taken of either side of the equal.

Exercise 9: Integer Powers and Polar Form#

Question a#

Write \(-1+\sqrt{3}i\) in polar form and use it in a similar manner as in Example 4.6.2 to show that

\[(-1+\sqrt{3}i)^{10}=2^{9}(-1+\sqrt{3}i).\]

Question b#

Let \(n\) be a natural number. Show the following:

\[(-1+\sqrt{3}i)^{3n}=2^{3n},\]
\[(-1+\sqrt{3}i)^{3n+1}=2^{3n}(-1+\sqrt{3}i),\]

and

\[(-1+\sqrt{3}i)^{3n+2}=2^{3n+1}(-1-\sqrt{3}i).\]

Hint

First, show that \((-1+\sqrt{3}i)^{3}=2^3\). What can now be said about \((-1+\sqrt{3}i)^{3n}\)?

Hint

If \((-1+\sqrt{3}i)^{3}=2^3\), then \(((-1+\sqrt{3}i)^{3})^n=(2^3)^n\,\), \((-1+\sqrt{3}i)^{3n+1}=(2^3)^n(-1+\sqrt{3}i)\,\), and \((-1+\sqrt{3}i)^{3n+2}=(2^3)^n(-1+\sqrt{3}i)^2\).

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