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Complex Analysis: an Open Source Textbook

Russell W. Howell, John H. Mathews

Section 1.2 The Algebra of Complex Numbers, Part I

We have shown that complex numbers came to be viewed as ordered pairs of real numbers. That is, a complex number \(z\) is defined to be
\begin{equation} z=(x, y)\text{,}\tag{1.2.1} \end{equation}
where \(x\) and \(y\) are both real numbers.
The reason we say ordered pair is because we are thinking of a point in the plane. The point \((2, 3)\text{,}\) for example, is not the same as \((3, 2)\text{.}\) The order in which we write \(x\) and \(y\) in Equation (1.2.1) makes a difference. Clearly, then, two complex numbers are equal if and only if their \(x\) coordinates are equal and their \(y\) coordinates are equal. In other words,
\begin{equation*} (x, y)=(u, v) \quad \text{iff} \quad x=u \quad \text{and} \quad y=v\text{.} \end{equation*}
(Throughout this text, “iff” means if and only if.)
A meaningful number system requires a method for combining ordered pairs. The definition of algebraic operations must be consistent so that the sum, difference, product, and quotient of any two ordered pairs will again be an ordered pair. The key to defining how these numbers should be manipulated is to follow Gauss’s lead and equate \((x, y)\) with \(x+iy\text{.}\) Then, if \(z_1 = (x_1, y_1)\) and \(z_2 = (x_2, y_2)\) are arbitrary complex numbers, we have
\begin{align*} z_1+z_2 \amp = (x_1, y_1) + (x_2, y_2)\\ \amp =(x_1+iy_1) + (x_2+iy_2)\\ \amp =(x_1 +x_2) +i(y_1 +y_2)\\ \amp =(x_1 + x_2,\ y_1 +y_2)\\ \amp =(x_1 +x_2) +i(y_1 + y_2)\text{.} \end{align*}
Thus, the following definitions should make sense.

Definition 1.2.1. Addition.

\begin{align*} z_1 +z_2 \amp = (x_1, y_1) + (x_2, y_2)\\ \amp =(x_1 + x_2, \, y_1 +y_2) \text{.} \end{align*}

Definition 1.2.2. Subtraction.

\begin{align*} z_1 - z_2 \amp =(x_1, y_1) - (x_2, y_2)\\ \amp =(x_1-x_2, \, y_1-y_2) \text{.} \end{align*}

Example 1.2.3.

If \(z_1=(3, 7)\) and \(z_2=(5, -6)\text{,}\) then
\begin{align*} z_1 + z_2 \amp =(3,7) + (5,-6) = (8, 1), \quad \textit{and}\\ z_1 - z_2 \amp =(3,7) - (5,-6) = (-2,13)\text{.} \end{align*}
We can also use the notation \(z_1=3+7i\) and \(z_2=5-6i\text{:}\)
\begin{align*} z_1 + z_2 \amp =(3+7i)+(5-6i)=8+i, \quad \textit{and}\\ z_1 - z_2 \amp =(3+7i)-(5-6i)=-2+13i\text{.} \end{align*}
Given the rationale we devised for addition and subtraction, it is tempting to define the product \(z_1z_2\) as \(z_1 z_2=(x_1 x_2, y_1 y_2)\) . It turns out, however, that this is not a good definition, and we ask you in the exercises for this section to explain why. How, then, should products be defined? Again, if we equate \((x,\ y)\) with \(x+iy\) and assume, for the moment, that \(i=\sqrt{-1}\) makes sense (so that \(i^2=-1\)), we have
\begin{align*} z_1z_2 \amp =(x_1, y_1)(x_2, y_2)\\ \amp = (x_1+iy_1)(x_2+iy_2)\\ \amp = x_1x_2 + ix_1y_2 + ix_2y_1 +i^2 y_1y_2\\ \amp = x_1x_2 - y_1y_2 +i(x_1y_2 + x_2y_1)\\ \amp = (x_1x_2 - y_1y_2, \, x_1y_2 + x_2y_1)\text{.} \end{align*}
It appears, therefore, that we are forced into the following definition.

Definition 1.2.4. Multiplication.

\begin{align} z_1z_2 \amp = (x_1, y_1)(x_2, y_2)\notag\\ \amp = (x_1x_2 - y_1y_2, \, x_1y_2 + x_2y_1)\text{.}\tag{1.2.2} \end{align}

Example 1.2.5.

If \(z_1=(3,\ 7)\) and \(z_2=5-6i\text{,}\) then
\begin{align*} z_1z_2 \amp = (3, 7)(5, -6)\\ \amp = (3 \cdot 5 - 7 \cdot (-6), \; 3 \cdot (-6) + 5 \cdot 7)\\ \amp =(15+42, \, -18+35)\\ \amp =(57, 17)\text{.} \end{align*}
We get the same answer by using the notation \(z_1=3+7i\) and \(z_2=5-6i\text{:}\)
\begin{align*} z_1z_2 \amp = (3, 7)(5, -6)\\ \amp = (3+7i)(5-6i)\\ \amp = 15-18i+35i-42i^2\\ \amp = 15-42(-1)+(-18+35)i\\ \amp =57+17i\\ \amp =(57, 17)\text{.} \end{align*}
Of course, it makes sense that the answer came out as we expected because we used the notation \(x+iy\) as motivation for our definition in the first place. Exercise 1.2.14 asks you to show that Wessel’s analogy for the norm and angular displacement discussed in Section 1.1.4 leads to the same rule for multiplication as that given in Definition 1.2.4.
To motivate our definition for division, we proceed along the same lines as we did for multiplication, assuming that \(z_2 \ne 0\text{:}\)
\begin{align*} \frac{z_1}{z_2} \amp =\frac{(x_1, y_1)}{(x_2, y_2)}\\ \amp = \frac{(x_1+iy_1)}{(x_2+iy_2)}\text{.} \end{align*}
We need to figure out a way to write the preceding quantity in the form \(x+iy\text{.}\) To do so, we use a standard technique and multiply the numerator and denominator by \(x_2-iy_2\text{,}\) which gives
\begin{align*} \frac{z_1}{z_2} \amp = \frac{(x_1+iy_1)(x_2-iy_2)}{(x_2+iy_2)(x_2-iy_2)}\\ \amp = \frac{x_1x_2+y_1y_2 + i(-x_1y_2 + x_2y_1)}{x_2^2 + y_2^2}\\ \amp = \left(\frac{x_1x_2 + y_1y_2}{x_2^2 + y_2^2}\right) + i\left(\frac{-x_1y_2 + x_2y_1}{x_2^2 + y_2^2}\right)\\ \amp = \left(\frac{x_1x_2 + y_1y_2}{x_2^2 + y_2^2}, \, \frac{-x_1y_2 + x_2y_1}{x_2^2 + y_2^2}\right)\text{.} \end{align*}
Thus we finally arrive at a rather odd definition.

Definition 1.2.6. Division.

\begin{align} \frac{z_1}{z_2} \amp = \frac{(x_1,y_1)}{(x_2,y_2)}\notag\\ \amp = \left(\frac{x_1x_2 + y_1y_2}{x_2^2 + y_2^2}, \, \frac {-x_1y_2 + x_2y_1}{x_2^2 + y_2^2}\right) \text{ for } z_2 \ne 0\text{.}\tag{1.2.3} \end{align}

Example 1.2.7.

If \(z_1=(3, 7)\) and \(z_2=(5, -6)\text{,}\) then
\begin{equation*} \frac{z_1}{z_2}=\frac{(3, \, 7)}{(5, \, -6)} = \left(\frac{15-42}{25+36}, \; \frac{18+35}{25+36}\right) = \left(-\frac{27}{61}, \; \frac{53}{61}\right)\text{.} \end{equation*}
As with the example for multiplication, we also get this answer if we use the notation \(x+iy\text{:}\)
\begin{align*} \frac{z_1}{z_2} \amp =\frac{(3, \, 7)}{(5, \, -6)}\\ \amp =\frac{3+7i}{5-6i}\\ \amp =\left(\frac{3+7i}{5-6i}\right) \left(\frac{5+6i}{5+6i}\right)\\ \amp =\frac{15+18i+35i+42i^2}{25+30i-30i-36i^2}\\ \amp =\frac{15-42+(18+35)i}{25+36}\\ \amp =-\frac{27}{61}+\frac{53}{61}i\\ \amp =\left(-\frac{27}{61}, \; \frac{53}{61}\right)\text{.} \end{align*}
To perform operations on complex numbers, mathematicians use the notation \(x+iy\) and engage in algebraic manipulations, as we did here, rather than apply the complicated-looking definitions we gave for those operations on ordered pairs. This procedure is valid because we used the \(x+iy\) notation as a guide for defining the operations in the first place. Remember, though, that the \(x+iy\) notation is nothing more than a convenient bookkeeping device for keeping track of how to manipulate ordered pairs. It is the ordered pair algebraic definitions that form the real foundation on which the complex number system is based. In fact, if you were to program a computer to do arithmetic on complex numbers, your program would perform calculations on ordered pairs, using exactly the definitions that we gave.
It turns out that our algebraic definitions give complex numbers all the properties we normally ascribe to the real number system. Taken together, they describe what algebraists call a field. In formal terms, a field is a set (in this case, the complex numbers) together with two binary operations (in this case, addition and multiplication) having the following properties.
  1. (P1) Commutative law for addition: \(z_1+z_2=z_2+z_1\text{.}\)
  2. (P2) Associative law for addition: \(z_1+(z_2+z_3)=(z_1+z_2)+z_3\text{.}\)
  3. (P3) Additive identity: There is a complex number \(\omega\) such that \(z+\omega=z\) for all complex numbers \(z\text{.}\) The number \(\omega\) is obviously the ordered pair \((0, 0)\text{.}\)
  4. (P4) Additive inverses: For any complex number \(z\text{,}\) there is a unique complex number \(\eta\) (depending on \(z\)) with the property that \(z+\eta =(0, 0)\text{.}\) Obviously, if \(z=(x, y)=x+iy\text{,}\) the number \(\eta\) will be \((-x, -y)=-x-iy=-z\text{.}\)
  5. (P5) Commutative law for multiplication: \(z_1z_2=z_2z_1\text{.}\)
  6. (P6) Associative law for multiplication: \(z_1(z_2z_3)=(z_1z_2)z_3\text{.}\)
  7. (P7) Multiplicative identity: There is a complex number \(\zeta\) such that \(z\zeta=z\) for all complex numbers \(z\text{.}\) As you might expect, \((1, 0)\) is the unique complex number \(\zeta\) having this property. We ask you to verify this identity in the exercises for this section.
  8. (P8) Multiplicative inverses: For any complex number \(z=(x,y)\) other than the number \((0, 0)\text{,}\) there is a complex number (depending on \(z\)), which we denote \(z^{-1}\text{,}\) having the property that \(zz^{-1}=(1, 0)=1\text{.}\) Based on our definition for division, it seems reasonable that the number \(z^{-1}\) would be
    \begin{align*} z^{-1}=\frac{(1,0)}{z} = \frac{1}{z}=\frac{1}{x+iy} = \frac{x-iy}{x^2 + y^2} \amp = \frac{x}{x^2+y^2}+i\left(\frac{-y}{x^2+y^2}\right)\\ \amp = \left(\frac{x}{x^2+y^2}, \, \frac{-y}{x^2+y^2}\right)\text{.} \end{align*}
    We ask you to confirm this result in the exercises for this section.
  9. (P9) The distributive law: \(z_1(z_2+z_3)=z_1z_2+z_1z_3\text{.}\)
None of these properties is difficult to prove. Most of the proofs make use of corresponding facts in the real number system. To illustrate, we give a proof of property (P1).
Proof of the commutative law for addition:
Let \(z_1=(x_1,\ y_1)\) and \(z_2=(x_2,\ y_2)\) be arbitrary complex numbers. Then
\begin{align*} z_1+z_2 \amp = (x_1, y_1) + (x_2, y_2) \amp \amp\\ \amp = (x_1+x_2, y_1+y_2) \amp \amp \text{ (by definition of addition of complex numbers) }\\ \amp = (x_2+x_1, y_2+y_1) \amp \amp \text{ by the commutative law for real numbers )}\\ \amp = (x_2, y_2)+(x_1, y_1) \amp \amp \text{ (by definition of addition of complex numbers) }\\ \amp = z_2+z_1\text{.} \end{align*}
Actually, you can think of the real number system as a subset of the complex number system. To see why, let’s agree that, as any complex number of the form \((t,\,0)\) is on the \(x\) axis, we can identify it with the real number \(t\text{.}\) With this correspondence, we can easily verify that our definitions for addition, subtraction, multiplication, and division of complex numbers are consistent with the corresponding operations on real numbers. For example, if \(x_1\) and \(x_2\) are real numbers, then
\begin{align*} x_1x_2 \amp = (x_1, 0)(x_2, 0) \amp \amp \text{ (by our agreed correspondence) }\\ \amp = (x_1x_2-0,\ 0+0) \amp \amp \text{ (by definition of multiplication of complex numbers) }\\ \amp = (x_1x_2, 0) \amp \amp \text{ (confirming the consistency of our correspondence). } \end{align*}
It is now time to show specifically how the symbol \(i\) relates to the quantity \(\sqrt{-1}\text{.}\) Note that
\begin{align*} (0, 1)^2 \amp = (0, 1)(0, 1) \amp \amp\\ \amp = (0-1, \, 0+0) \amp \amp \text{ (by definition of multiplication of complex numbers) }\\ \amp = (-1, 0) \amp \amp\\ \amp = -1 \amp \amp \text{ (by our agreed correspondence). } \end{align*}
If we use the symbol \(i\) for the point \((0, 1)\text{,}\) the preceding identity gives
\begin{equation*} i^2=(0, 1)^2=-1\text{,} \end{equation*}
which means \(i=(0,\ 1)=\sqrt{-1}\text{.}\) So, the next time you are having a discussion with your friends and they scoff when you claim that \(\sqrt{-1}\) is not imaginary, calmly put your pencil on the point \((0, 1)\) of the coordinate plane and ask them if there is anything imaginary about it. When they agree there isn’t, you can tell them that this point, in fact, represents the mysterious \(\sqrt{-1}\) in the same way that \((1, 0)\) represents the number 1.
We can also see more clearly now how the notation \(x+iy\) equates to \((x,\ y)\) . Using the preceding conventions (\ie, \(x=(x, 0)\text{,}\) etc.), we have
\begin{align*} x+iy \amp = (x, 0)+(0, 1)(y, 0) \amp \amp \text{ (by our previously discussed conventions) }\\ \amp = (x, 0)+(0, y) \amp \amp \text{ (by definition of multiplication of complex numbers) }\\ \amp = (x, y) \amp \amp \text{ (by definition of addition of complex numbers). } \end{align*}
Thus, we may move freely between the notations \(x+iy\) and \((x, y)\text{,}\) depending on which is more convenient for the context in which we are working. Students sometimes wonder whether it matters where the “\(i\)” is located in writing a complex number. It does not. Generally, most texts place terms containing an “\(i\)” at the end of an expression, and place the “\(i\)” before a variable, but after a constant. Thus, we write \(x+iy,\, u+iv\text{,}\) etc., but \(3+7i, \, 5-6i\) and so forth. Because letters lower in the alphabet generally denote constants, you will usually (but not always) see the expression \(a+bi\) instead of \(a+ib\text{.}\) Many authors write quantities like \(1+i\sqrt{3}\) instead of \(1+\sqrt{3}i\) to make sure the “\(i\)” is not mistakenly thought to be inside the square root symbol. Additionally, if there is concern that the “\(i\)” might be missed, it is sometimes placed before a lengthy expression, as in \(2\cos(-\frac{5\pi}{6}+2n\pi) +i2\sin(-\frac{5\pi}{6}+2n\pi)\text{.}\)
We close this section with three important definitions and a theorem involving them. We ask you for a proof of the theorem in the exercises.

Definition 1.2.8. Real Part.

The real part of \(z\text{,}\) denoted \(\mathrm{Re}(z)\text{,}\) is the real number \(x\text{.}\)

Definition 1.2.9. Imaginary Part.

The imaginary part of \(z\text{,}\) denoted \(\mathrm{Im}(z)\text{,}\) is the real number \(y\text{.}\)

Definition 1.2.10. Conjugate.

The conjugate of \(z\text{,}\) denoted \(\overline{z}\text{,}\) is the complex number \((x, -y)=x-iy\text{.}\)

Example 1.2.11.

\(\mathrm{Re}(-3+7i) =-3 \text{ and } \mathrm{Re}[(9, 4)]=9, \, \mathrm{Im}(-3+7i)=7\text{.}\) Also, \(\mathrm{Im}[(9, 4)]=4, \, \overline{-3+7i}=-3-7i \text{ and } \overline{(9,\ 4)}=(9, -4)\text{.}\)
Because of what it erroneously connotes, it is a shame that the term imaginary is used in Definition 1.2.9. It was coined by the brilliant mathematician and philosopher René Descartes (1596–1650) during an era when quantities such as \(\sqrt{-1}\) were thought to be just that. Gauss, who was successful in getting mathematicians to adopt the phrase complex number rather than imaginary number, also suggested that we use lateral part of \(z\) in place of imaginary part of \(z\text{.}\) Unfortunately, that suggestion never caught on, and it appears we are stuck with what history has handed down to us.

Exercises Exercises

1.

Perform the required calculations and express your answers in the form \(a+bi\text{.}\)
(a)
\(i^{275}\text{.}\)
Solution.
\(i^{275}=(i^2)^{137}i = (-1)^{137}i = -i\text{.}\)
(b)
\(\frac{1}{i^{5}}\text{.}\)
(c)
\(\mathrm{Re}(i)\text{.}\)
Solution.
\(0\text{.}\)
(d)
\(\mathrm{Im}(2)\text{.}\)
(e)
\((i-1)^3\text{.}\)
Solution.
\(2+2i\text{.}\)
(f)
\((7-2i)(3i+5)\text{.}\)
(g)
\(\mathrm{Re}(7+6i) +\mathrm{Im}(5-4i). \label {1.2.1g}\)
Solution.
\(3\text{.}\)
(h)
\(\mathrm{Im}(\frac{1+2i}{3-4i})\text{.}\)
(i)
\(\frac{(4-i)(1-3i)}{-1+2i}\text{.}\)
Solution.
\(-\frac{27}{5}+\frac{11}{5}i\text{.}\)
(j)
\(\overline{(1+i\sqrt{3})(i+\sqrt{3})}\text{.}\)

2.

Evaluate the following quantities.
(a)
\(\overline{(1+i)(2+i)}(3+i)\text{.}\)
(b)
\((3+i)/(\overline{2+i})\text{.}\)
(c)
\(\mathrm{Re}(i-1)^3] \text{.}\)
(d)
\(\mathrm{Im}[(1+i)^-2]\text{.}\)
(e)
\(\frac{1+2i}{3-4i}-\frac{4-3i}{2-i}\text{.}\)
(f)
\((1+i)^{-2}\text{.}\)
(g)
\(\mathrm{Re}[(x-iy)^2]\text{.}\)
(h)
\(\mathrm{Im}(\frac{1}{x-iy})\text{.}\)
(i)
\(\mathrm{Re}[(x+iy)(x-iy)]\text{.}\)
(j)
\(\mathrm{Im}[(x+iy)^3]\text{.}\)

3.

Show that \(z\overline{z}\) is always a real number.
Solution.
Let \(z=x+iy\) be an arbitrary complex number. Then \(z \overline{z}=(x+iy)(x-iy)=x^2+y^2\text{,}\) which is obviously a real number.

5.

Let \(P(z)=a_nz^n + a_{n-1}z^{n-1} + \cdots +a_1z + a_0\) be a polynomial of degree \(n\text{.}\)
(a)
Suppose that \(a_n, a_{n-1}, \ldots, a_1, a_0\) are all real. Show that if \(z_1\) is a root of \(P\text{,}\) then \(\overline{z_1}\) is also a root. In other words, the roots must be complex conjugates, something you likely learned without proof in high school.
Solution.
Since \(z_1\) is a root of the polynomial \(P\text{,}\) \(P(z_1)=0\text{.}\) Use properties (1.2.4) through (1.2.6) of Theorem 1.2.12 to show that \(\overline{P\left(\overline{z_1}\right)} =P(z_1)\text{,}\) which implies \(\overline{P\left(\overline{z_1}\right)}=0\text{.}\) Next show that if \(\overline{P\left(\overline{z_1}\right)}=0\text{,}\) then \(P\left(\overline{z_1}\right)=0\text{,}\) confirming that \(\overline{z_1}\) is also a root of \(P\text{.}\)
(b)
Suppose not all of \(a_n, a_{n-1}, \ldots, a_1, a_0\) are real. Show that \(P\) has at least one root whose complex conjugate is not a root. \hint{Prove the contrapositive.}
(c)
Find an example of a polynomial that has some roots occurring as complex conjugates, and some not.
Solution.
Find a polynomial for part a, another for part b, and multiply them together.

6.

Let \(z_1=(x_1, y_1)\) and \(z_2=(x_2, y_2)\) be arbitrary complex numbers. Prove or disprove the following.
(a)
\(\mathrm{Re}(z_1+z_2)=\mathrm{Re}(z_1)+\mathrm{Re}(z_2)\text{.}\)
(b)
\(\mathrm{Re}(z_1z_2)=\mathrm{Re}(z_1)\mathrm{Re}(z_2) \text{.}\)
(c)
\(\mathrm{Im}(z_1+z_2) =\mathrm{Im}(z_1) + \mathrm{Im}(z_2)\text{.}\)
(d)
\(\mathrm{Im}(z_1z_2)=\mathrm{Im}(z_1)\mathrm{Im}(z_2)\text{.}\)

7.

Prove that the complex number \((1, 0)\) (which we identify with the real number 1) is the multiplicative identity for complex numbers.
Solution.
Use the (ordered pair) definition for multiplication to verify that if \(z=(x,y)\) is any complex number, then \((x,y)(1,0)=(x,y)\text{.}\)

8.

Use mathematical induction to show that the binomial theorem is valid for complex numbers. In other words, show that, if \(z\) and \(w\) are arbitrary complex numbers and \(n\) is a positive integer, then
\begin{equation} (z+w)^n=\sum\limits_{k=0}^n\binom{n}{k}z^kw^{n-k}, \text{ where } \binom{n}{k}= \frac{n!}{k!(n-k)!}\text{.}\tag{1.2.10} \end{equation}

9.

Let’s use the symbol \(\ast\) for a new type of multiplication of complex numbers defined by \(z_1\ast z_2=(x_1x_2,\ y_1y_2)\text{.}\) This exercise shows why this is a bad definition.
(a)
Use the definition given in property (P7) and state what the multiplicative identity \(\zeta\) would have to be for this new multiplication.
Solution.
We would want to find a number \(\zeta =(a,b)\) such that for any \(z=(x,y)\) we have \(z\ast \zeta =z\text{.}\) Obviously, if \(\zeta =(1,1)\text{,}\) then according to the definition of \(\ast\) we would have \(z\ast \zeta =(x,y)\ast(1,1)=(x,y)=z\text{.}\) Thus, the multiplicative identity in this case would have to be \(\zeta =(1,1)\text{.}\)
(b)
Show that, if you use this new multiplication, nonzero complex numbers of the form \((0, a)\) have no inverse. That is, show that, if \(z=(0,\,a)\text{,}\) there is no complex number \(w\) with the property that \(z\ast w=\zeta\text{,}\) where \(\zeta\) is the multiplicative identity you found in part (a). \label {1.2.9b}
Solution.
For any complex number \(w=(x,y)\) we would have \((0,a \ast (x,y)=(0,a)\text{,}\) which can’t possibly equal \((1,1)\text{.}\)

10.

Explain why the complex number \((0, 0)\) (which, you recall, we identify with the real number 0) has no multiplicative inverse.

11.

Prove property (P9), the distributive law for complex numbers.
Solution.
Let \(z_1=(x_1,y_1), \, z_2=(x_2,y_2)\text{,}\) and \(z_3=(x_3,y_3)\) be arbitrary complex numbers. Then
\begin{align*} z_1(z_2+z_3) \amp = (x_1,\,y_1)\big[(x_2,\,y_2)+(x_3,\,y_3)\big]\\ \amp =(x_1,\,y_1)\big[(x_2+x_3,\,y_2+y_3)\big]\\ \amp =\big(x_1(x_2+x_3)-y_1(y_2+y_3),\,x_1(y_2+y_3)+(x_2+x_3)y_1\big)\\ \amp =\cdots\\ \amp = (x_1x_2-y_1y_2, \, x_1y_2+x_2y_1) + (x_1x_3-y_1y_3, \, x_1y_3+x_2y_3)\\ \amp = z_1z_2 + z_1z_3\text{.} \end{align*}
Complete the missing steps in \(\cdots\) above by using algebraic properties of real numbers.

12.

Verify that, if \(z=(x, y)\text{,}\) with \(x\) and \(y\) not both 0, then \(z^{-1}= \frac{{(1, 0)}}{z} \; \big(\text{ i.e., } z^{-1}=\frac{{1}}{z}\big)\text{.}\)
Hint.
\(z=(x, y)\)\(z^{-1}=\frac{(1, 0)}{(x, y)}\text{.}\)\(zz^{-1}=(1, 0)=1\text{.}\)

13.

From Exercise 1.2.12 and basic cancellation laws, it follows that \(z^{-1}=\frac{1}{z}=\frac{\overline{z}}{z\overline{z}}\text{.}\) The numerator here, \(\overline{z}\text{,}\) is trivial to calculate and, as the denominator \(z \overline{z}\) is a real number (Exercise 1.2.3), computing the quotient \(\frac{\overline{z}}{z\overline{z}}\) should be rather straightforward. Use this fact to compute \(z^{-1}\) if \(z=2+3i\) and again if \(z=7-5i\text{.}\)
Solution.
\((2+3i)^{-1}=\frac{2}{13}-\frac{3}{13}i, (7-5i)^{-1}=\frac{7}{74}+\frac{5}{74}i\text{.}\)

14.

Recall the following trigonometric identities:
(a)
\(\cos(\theta_1+\theta_2) = \cos\theta_1 \cos\theta_2 - \sin\theta_1 \sin\theta_2\text{;}\)
(b)
\(\sin(\theta_1+\theta_2) = \cos\theta_1 \sin\theta_2 + \sin\theta_1 \cos\theta_2\text{.}\)
Use these identities to show (using Wessel’s analogy for the norm and angular displacement discussed in Section 1.1.4) that the ordered-pair definition for the product of two vectors must agree with Equation (1.2.2) of the text. In other words, show that it must be the case that, if \(\mathbf{z_1}= (x_1,y_1)\) and \(\mathbf{z_2}=(x_2,y_2)\text{,}\) then \(\mathbf{z_1z_2}=(x_1x_2-y_1y_1,\,x_1y_2+x_2y_1)\text{.}\)
\hint{It may be helpful first to assume that both vectors are unit vectors, \ie, that they are somewhere on the unit circle.}

15.

Show, by equating the real numbers \(x_1\) and \(x_2\) with \((x_1, 0)\) and \((x_2, 0)\text{,}\) respectively, that the complex definition for division is consistent with the real definition for division.
Hint.
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