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+---
+title: Complex Numbers
+---
+
+Complex numbers
+===============
+
+Complex numbers are numbers of the form $$z = a + b {\rm i}.$$ Here
+$\rm i$ is the square root of -1: $${\rm i} = \sqrt{-1},$$ or,
+equivalently: $${\rm i}^2 = -1.$$
+
+Usual operations for numbers have their natural extension for complex
+numbers as we shall see below.
+
+Some definitions:
+
+-   For a complex number $z = a + b {{\rm i}}$, $a$ is called the *real
+    part*, and $b$ the *imaginary part*.
+
+-   The *complex conjugate* $z^*$ of $z = a + b {{\rm i}}$ is defined as
+    $$z^* = a - b{{\rm i}},$$ i.e., taking the complex conjugate means
+    flipping the sign of the imaginary part.
+
+Addition
+--------
+
+For two complex numbers, $z_1 = a_1 + b_1 {{\rm i}}$ and
+$z_2 = a_2 + b_2 {{\rm i}}$, the sum $w = z_1 + z_2$ is given as
+$$w = w_1 + w_2 {{\rm i}}= (a_1 + a_2) + (b_1 + b_2) {{\rm i}}$$ where
+the parentheses in the rightmost expression have been added to group the
+real and the imaginary part. A consequence of this definition is that
+the sum of a complex number and its complex conjugate is real:
+$$z + z^* = a + b {{\rm i}}+ a - b {{\rm i}}= 2a,$$ i.e., this results
+in twice the real part of $z$. Similarly, subtracting $z^*$ from $z$
+yields $$z - z^* = a + b {{\rm i}} - a + b {{\rm i}}= 2b{\rm i},$$ i.e.,
+twice the imaginary part of $z$ (times $\rm i$).
+
+Multiplication
+--------------
+
+For the same two complex numbers $z_1$ and $z_2$ as above, their product
+is calculated as
+$$w = z_1 z_2 = (a_1 + b_1 {{\rm i}}) (a_2 + b_2 {{\rm i}}) = (a_1 a_2 - b_1 b_2) + (a_1 b_2 + a_2 b_1) {{\rm i}},$$
+where the parentheses have again be used to indicate the real and
+imaginary parts.
+
+A consequence of this definition is that the product of a complex number
+$z = a + b {{\rm i}}$ with its conjugate is real:
+$$z z^* = (a+b{{\rm i}})(a-b{{\rm i}}) = a^2 + b^2.$$ The square root of
+this number is the *norm* $|z|$ of $z$:
+$$|z| = \sqrt{z z^*} = \sqrt{a^2 + b^2}.$$
+
+Division
+--------
+
+The quotient $z_1/z_2$ of two complex numbers $z_1$ and $z_2$ as above,
+can be evaluated by multiplying the numerator and denominator by the
+complex conjugate of $z_2$:
+$$\frac{z_1}{z_2} = \frac{z_1 z_2^*}{z_2 z_2^*} = \frac{(a_1 a_2 + b_1 b_2) + (-a_1 b_2 + a_2 b_1) {{\rm i}}}{a_2^2 + b_2^2}.$$
+Check this!
+
+The complex plane
+=================
+
+Complex numbers can be rendered on a two-dimensional (2D) plane, the
+*complex plane*. This plane is spanned by two unit vectors, one
+horizontal, which represents the real number 1, whereas the vertical
+unit vector represents ${\rm i}$.
+
+![image](complex_numbers_files/complex_numbers_5_0.pdf)
+
+[\
+]{}
+
+Note that the norm of $z$ is the length of this vector.
+
+Addition in the complex plane
+-----------------------------
+
+Adding two numbers in the complex plane corresponds to adding the
+horizontal and vertical components:
+
+![image](complex_numbers_files/complex_numbers_8_0.pdf)
+
+[\
+]{}
+
+We see that the sum is found as the diagonal of a parallelogram spanned
+by the two numbers.
+
+Argument and Norm
+-----------------
+
+A complex number can be represented by two real numbers, $a$ and $b$
+which represent the real and imaginary part of the complex number. An
+alternative representation is a *vector* in the complex plane, whose
+horizontal component is the real, and vertical component the imaginary
+part. However, it is also possible to characterize that vector by its
+*length* and *direction*, where the latter can be represented by the
+angle the vector makes with the horizontal axis:
+
+![image](complex_numbers_files/complex_numbers_10_0.pdf)
+
+[\
+]{}
+
+The angle with the horizontal axis is denoted by $\varphi$, just as in
+the case of polar coordinates. In the context of complex numbers, this
+angle is denoted as the *argument*. We have:
+
+> A complex number can be represented either by its real and imaginary
+> part, corresponding to the Cartesian coordinates in the complex plane,
+> or by its *norm* and its *argument*, corresponding to polar
+> coordinates. The norm is the length of the vector, and the argument is
+> the angle it makes with the horizontal axis.
+
+From our previous discussion on polar coordinates we can conclude that
+for a complex number $z = a + b {\rm i}$, its real and imaginary parts
+can be expressed as $$a = |z| \cos\varphi$$ $$b = |z| \sin\varphi$$ The
+inverse equations are $$|z| = \sqrt{a^2 + b^2}$$
+$$\varphi = \arctan(b/a)$$ for $a>0$. In general:
+$$\varphi = \begin{cases} \arctan(b/a) &{\rm for ~} a>0; \\
+ \pi + \arctan(b/a) & {\rm for ~} a<0 {\rm ~ and ~} b>0;\\
+ -\pi + \arctan(b/a) &{\rm ~ for ~} a<0 {\rm ~ and ~} b<0.
+ \end{cases}$$
+
+Complex functions
+-----------------
+
+A complex function $f$ maps any complex number $z$ onto another complex
+number $f(z)$. Just as with real functions, we can define the derivative
+of a complex function:
+$$\frac{df(z)}{dz} = \lim_{dz \rightarrow 0} \frac{f(z+dz) - f(z)}{dz}.$$
+It is useful to contemplate this expression for some time. The number
+$dz$ in this expression is a *complex* number with vanishingly small
+norm. However, its *direction* (i.e. its *argument*) is not specified. A
+complex function is called *differentiable* only if the limit exists and
+gives one and the same number, irrespective of the direction of $dz$.
+
+Let’s see whether we can find a condition for this to happen. We assume
+that the function $f$ is continuous. Note that, as $f$ is complex, it
+can be written as a real plus an imaginary part:
+$$f(z) = u(z) + {\rm i} v(z).$$ For $z=x + \rm i y$, we can write the
+two functions $u$ and $v$ in terms of $x$ and $y$. Let us assume that
+the two partial derivatives of these functions with respect to $x$ and
+$y$ exist.
+
+Calculating the partial derivatives of $f$ with respect to $x$ and $y$
+yields:
+$$\frac{\partial f(z)}{\partial x} = \lim_{dx \rightarrow 0} \frac{f(z+dx) - f(z)}{dx} = \frac{\partial u}{\partial x} + {\rm i} \frac{\partial v}{\partial x},$$
+and
+$$\frac{\partial f(z)}{\partial y} = \lim_{dy \rightarrow 0} \frac{f(z+{\rm i} dy) - f(z)}{dy} = \frac{\partial u}{\partial y} + {\rm i} \frac{\partial v}{\partial y}.$$
+Unlike $df/dz$, these derivatives are always defined. In terms of these
+two partial derivatives, we can find the derivative with respect to any
+direction using first-order Taylor expansions. Taking
+$dz = d x + {\rm i} dy$, we have
+$$f(z + dz) = u(x+dx, y+dy) + {\rm i} v(x+dx, y+dy) = u(x,y) + {\rm i} v(x,y) + dx \frac{\partial u}{\partial x} + dy \frac{\partial u}{\partial y} + 
+{\rm i} \left[ dx \frac{\partial v}{\partial x} + dy \frac{\partial  v}{\partial y} \right].$$
+Collecting the terms proportional to $dx$ and $dy$ respectively we
+obtain:
+$$f(z + dz ) = f(z) + \left( \frac{\partial u}{\partial x} + {\rm i} \frac{\partial v}{\partial x} \right) dx +
+\left( \frac{\partial u}{\partial y} + {\rm i} \frac{\partial v}{\partial y} \right) dy.$$
+Let’s write
+$$f(z + dz ) - f(z) = \left( \frac{\partial u}{\partial x} + {\rm i}\frac{\partial v}{\partial x} \right) dx +
+{\rm i} \left( -{\rm i} \frac{\partial u}{\partial y} + \frac{\partial v}{\partial y} \right) dy.$$
+If the derivative $df/dz$ exists, the right hand side divided by $dz$
+should be independent of $dz=dx + {\rm i} dy$! Thus, $f(z)$ is
+differentiable only when
+$$\frac{\partial u}{\partial x} + {\rm i} \frac{\partial v}{\partial x} = -{\rm i} \frac{\partial u}{\partial y} + \frac{\partial v}{\partial y}.$$
+Equating the real and imaginary parts of the left and right hand side we
+obtain the
+
+TODO: Here was a remark environment
+
+[ *Cauchy Riemann* differential equations:
+$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} {~~~ \rm and ~~~ } \frac{\partial v}{\partial x} = - 
+\frac{\partial u}{\partial y}.$$ The derivative is then given as
+$$\frac{df}{dz} = \frac{\partial u}{\partial x} + {\rm i} \frac{\partial v}{\partial x}.$$
+A complex function whose real and imaginary part ($u$ and $v$) obey the
+Cauchy-Riemann differential equations in a point $z$, is complex
+differentiable at the point $z$. ]{}
+
+Note that differentiability is a property which not only pertains to a
+function, but also to a point.
+
+It can be shown that a complex function which is differentiable in $z$,
+is infinitely differentiable. Such a function is called *analytic* in
+the point $z$. An analytic function can be expanded as an inifinite
+series near any point where the function is analytic. Here, “near any”
+point indicates that the series converges and coincides with $f$ only on
+a *circle* in the complex plane. The radius of that circle is limited to
+the distance of $z$ to the nearest point $w$ where the function is not
+analytic. This distance is called the *convergence radius*.
+
+Recall that the Taylor series around a complex point $a$ has the form
+$$f(z) = \sum_{j=0}^\infty d_j (z-a)^j,$$ where
+$$d_j = \frac{1}{j!} f^{(j)} (a).$$ Here, $f^{(j)}(a)$ denotes the
+$j$-th derivative of the function $f$ in the point $a$.
+
+The complex exponential function
+--------------------------------
+
+The exponential function $f(z) = \exp(z) = e^z$ is defined as:
+$$\exp(z) = e^{x} \left( \cos y + {\rm i} \sin y\right).$$
+
+**Exercise** Check that this function obeys
+$$\exp(z_1) \exp(z_2) = \exp(z_1 + z_2).$$ You need sum- and difference
+formulas of cosine and sine.
+
+**Exercise** Check that $\exp(z)$ obeys the Cauchy-Riemann equations and
+that the derivative is the exponential function itself:
+$$\frac{d \exp(z)}{dz} = \exp(z).$$ Note that, for any complex number
+$z$, we can write $$z = |z| e^{\rm i \varphi},$$ where
+$\varphi = \text{arg}(z)$.
+
+In *real calculus*, the logarithmic function is the inverse of the
+exponential function. Similarly, we want the complex logarithm to be the
+inverse of the complex exponential function. Let’s write
+$w = \exp(z) = e^x(\cos y + \rm i \sin y)$. We know then that
+$\log(w) = z = x + \rm i y$. Realising that the norm of
+$\cos y + \rm i \sin y$ is 1 (check this!), we see that
+$\left|w\right| = e^x$. Therefore, the real part of $\log w$ is the real
+logarithm of $|w|$.
+
+The imaginary part of the $\log w$ should be $y$. Now, $y$ is the
+*argument* of $w$. All in all, we therefore see that
+$$\log w = \log|w| + \rm i \arg(w).$$
+
+The complex exponential is used *extremely often*. It occurs in Fourier
+transforms and is very convenient for doing calculations involving
+cosines and sines.
+
+Let us show some tricks where the simple properties of the exponential
+function helps in re-deriving trigonometric identities.
+
+1.  Take $|z_1| = |z_2| = 1$, and $\arg{(z_1)} = \varphi_1$ and
+    $\arg{(z_2)} = \varphi_2$. Then it is easy to see that
+    $z_i = \exp({\rm i} \varphi_i)$, $i=1, 2$. Then:
+    $$z_1 z_2 = \exp[{\rm i} (\varphi_1 + \varphi_2)].$$ The left hand
+    side can be written as
+    $$z_1 z_2 = \left[ \cos(\varphi_1) + {\rm i} \sin(\varphi_1) \right] \left[ \cos(\varphi_2) + {\rm i} \sin(\varphi_2) \right] = \cos\varphi_1 \cos\varphi_2 - \sin\varphi_1 \sin\varphi_2 + {\rm i} \left( \cos\varphi_1 \sin\varphi_2 + 
+    \sin\varphi_1 \cos\varphi_2 \right).$$ On the other hand, the right
+    hand side can be written as
+    $$\exp[{\rm i} (\varphi_1 + \varphi_2)] = \cos(\varphi_1 + \varphi_2) + {\rm i} \sin(\varphi_1 + \varphi_2).$$
+    Comparing the two expressions, equating their real and imaginary
+    parts, we find
+    $$\cos(\varphi_1 + \varphi_2) = \cos\varphi_1 \cos\varphi_2 - \sin\varphi_1 \sin\varphi_2;$$
+    $$\sin(\varphi_1 + \varphi_2) = \cos\varphi_1 \sin\varphi_2 + 
+    \sin\varphi_1 \cos\varphi_2.$$ Note that we used the resulting
+    formulas already in order to derive the properties of the
+    exponential function. The point is that you can use the properties
+    of the complex exponential to quickly find the form of gonometric
+    formulas which you easily forget.
+
+2.  As a final example, consider what we can learn from the derivative
+    of the exponential function:
+    $$\frac{d}{d\varphi} \exp({\rm i} \varphi) = {\rm i} \exp({\rm i} \varphi) .$$
+    Writing out the exponential in terms of cosine and sine, we see that
+    $$\cos'\varphi + {\rm i} \sin'\varphi = {\rm i} \cos\varphi - \sin\varphi.$$
+    where the prime $'$ denotes the derivative as usual. Equating real
+    and imaginary parts leads to $$\cos'\varphi = - \sin\varphi;$$
+    $$\sin'\varphi = \cos\varphi.$$
+
+Hyperbolic functions
+--------------------
+
+From
+$e^{\rm i \varphi} = \left( \cos\varphi + {\rm i} \sin\varphi\right)$,
+it immediately follows that
+$$\cos\varphi = \frac{e^{{\rm i} \varphi} + e^{-{\rm i} \varphi}}{2}.$$
+and
+$$\sin\varphi = \frac{e^{{\rm i} \varphi} - e^{-{\rm i} \varphi}}{2{\rm i}}.$$
+It is then tempting to generalise these functions for imaginary angles.
+These functions are known as hyperbolic functions. They are are called
+the hyperbolic cosine and hyperbolic sine functions and they are denoted
+as $\sinh$ and $\cosh$: $$\cosh(x) = \frac{e^x + e^{-x}}{2};$$
+$$\sinh(x) = \frac{e^x - e^{-x}}{2}.$$ From these definitions the
+following properties can easily be derived.
+
+1.  Derivatives $$\frac{d\cosh(x)}{dx} = \sinh(x);$$
+    $$\frac{d\sinh(x)}{dx} = \cosh(x).$$
+
+2.  $$\cosh^2(x) - \sinh^2(x) = 1.$$
+
+3.  ‘Double angle’ formulas: $$\cosh(2x) = \cosh^2(x) + \sinh^2(x);$$
+    $$\sinh(2x) = 2\cosh(x) \sinh(x).$$
+
+It may seem that these function are rather exotic; however they occur in
+everyday life: the shapes of power lines and of soap films can be
+described by hyperbolic cosines and sines!
+
+Finally, the hyperbolic tangent is defined as
+$$\tanh(x) = \frac{\sinh(x)}{\cosh(x)}.$$ Its derivative is given as
+$$\tanh'(x) = 1 + \frac{\sinh^2 x}{\cosh^2 (x)} = - \frac{1}{\cosh^2(x)}.$$
+
+Summary
+=======
+
+TODO: Here was the beginning of a mdframed env
+
+-   A complex number $z$ has the form $$z = a + b \rm i$$ where $a$ and
+    $b$ are both real, and $\rm i^2 = 1$. The real number $a$ is called
+    the *real part* of $z$ and $b$ is the *imaginary part*. Two complex
+    numbers can be added, subtracted and multiplied straightforwardly.
+    The quotient of two complex numbers $z_1=a_1 + \rm i b_1$ and
+    $z_2=a_2 + \rm i b_2$ is
+    $$\frac{z_1}{z_2} = \frac{z_1 z_2^*}{z_2 z_2^*} = \frac{(a_1 a_2 + b_1 b_2) + (-a_1 b_2 + a_2 b_1) {{\rm i}}}{a_2^2 + b_2^2}.$$
+
+-   Complex numbers can also be characterised by their *norm*
+    $|z|=\sqrt{a^2+b^2}$ and *argument* $\phi$. These coordinates
+    correspond to polar coordinates in the complex plane. For a complex
+    number $z = a + b {\rm i}$, its real and imaginary parts can be
+    expressed as $$a = |z| \cos\varphi$$ $$b = |z| \sin\varphi$$ The
+    inverse equations are $$|z| = \sqrt{a^2 + b^2}$$
+    $$\varphi = \begin{cases} \arctan(b/a) &{\rm for ~} a>0; \\
+     \pi + \arctan(b/a) & {\rm for ~} a<0 {\rm ~ and ~} b>0;\\
+     -\pi + \arctan(b/a) &{\rm ~ for ~} a<0 {\rm ~ and ~} b<0.
+     \end{cases}$$
+
+-   The derivative of a complex function $f(z)$ is defined as
+    $$\frac{df(z)}{dz} = \lim_{dz \rightarrow 0} \frac{f(z+dz) - f(z)}{dz}.$$
+    The right hand side depends on the direction of $dz$ in the complex
+    plane. The function is said to be *differentiable* if the right hand
+    side gives a unique value. This is the case when the real part $u$
+    and imaginary part $v$ of the function $f$ satisfy the
+    *Cauchy–Riemann* equations:
+    $$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} {~~~ \rm and ~~~ } \frac{\partial v}{\partial x} = - 
+    \frac{\partial u}{\partial y}.$$ A function which is differentiable,
+    is differentiable infinitely often. Such a function can be expanded
+    as a Taylor series:
+    $$f(z) = \sum_{j=0}^\infty \frac{1}{j!} f^{(j)} (a) (z-a)^j,$$ where
+    $f^{(j)} (a)$ is the $j$-th derivative of the function $f$ in $a$.
+
+-   Examples of differentiable functions:
+
+    -   The complex exponential:
+        $$e^z = e^{x} \left( \cos y + \rm i \sin y\right).$$
+
+    -   The complex logarithm: $$\log(z) = \log|z| +\rm i \arg(z).$$
+
+    -   The complex sine and cosine functions are defined as
+        $$\sin(z) = \frac{e^{\rm i z} - e^{-\rm i z}}{2\rm i}; \phantom{xxx} \cos(z) = \frac{e^{\rm i z} + e^{-\rm i z}}{2}.$$
+        The complex tangent is defined as $\tan(z) = \sin(z)/\cos(z)$.
+
+    -   Hyperbolic functions are defined as:
+        $$\sinh(z) = \frac{e^{z} - e^{-z}}{2}; \phantom{xxx} \cosh(z) = \frac{e^{z} + e^{-z}}{2}.$$
+
+TODO: Here was the end of a mdframed env
+
+Problems
+========
+
+1.  [\[]{}D1[\]]{} Given $a=1+2\rm i$ and $b=4-2\rm i$, draw in the
+    complex plane the numbers $a+b$, $a-b$, $ab$, $a/b$, $e^a$ and
+    $\ln(a)$.
+
+2.  [\[]{}D1[\]]{} Evaluate (i) $\rm i^{1/4}$, (ii)
+    $\left(-1+\rm i \sqrt{3}\right)^{1/2}$, (iii) $\exp(2\rm i^3)$.
+
+3.  [\[]{}D1[\]]{} Evaluate $$\left| \frac{a+b\rm i}{a-b\rm i} \right|$$
+    for real $a$ and $b$.
+
+4.  [\[]{}D1[\]]{} Show that $\cos x = \cosh(\rm i x)$ and
+    $\cos(\rm i x) = \cosh x$. Derive similar relations for $\sinh$ and
+    $\sin$.
+
+    Show that $\cosh^2 x - \sinh^2 x = 1$.
+
+    Also show that $\cosh x$ is a solution to the differential equation
+    $$y'' = \sqrt{1 + y'^2}.$$
+
+5.  [\[]{}D1[\]]{} Calculate the real part of
+    $\int_0^\infty e^{-\gamma t  +\rm i \omega t} dt$ ($\omega$ and
+    $\gamma$ are real; $\gamma$ is positive).
+
+6.  [\[]{}D1[\]]{} Is the function $f(z) = |z| = \sqrt{x^2 + y^2}$
+    analytic on the complex plane or not? If not, where is the function
+    not analytic?
+
+7.  [\[]{}D1[\]]{} Show that the Cauchy-Riemann equations imply that the
+    real and imaginary part of a differentiable complex function both
+    represent solutions to the Laplace equation, i.e.
+    $$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0,$$
+    for the real part $u$ of the function, and similarly for the
+    imaginary part $v$.
+
+8.  [\[]{}D3[\]]{} Show that the set of points $z$ obeying
+    $$| z - \rm i a| = \lambda |z + \rm i a|,$$ with $a$ and $\lambda$
+    real, form a circle with radius $2|\lambda/(1-\lambda^2) a|$
+    centered on the point $\rm i a (1+\lambda^2)/(1-\lambda^2)$,
+    provided $\lambda \neq 1$. What is the set like for $\lambda = 1$?
+
+9.  [\[]{}D2[\]]{} In two dimensions, the Coulomb potential is
+    proportional to $\log |r|$. Viewing the 2D space as a complex plane,
+    this is $\log |z|$. Consider a system consisting of charges $q_i$
+    placed at ‘positions’ $z_i$, all close to the origin. The point $z$
+    is however located far away from the origin. Use the Taylor
+    expansions of the terms $\ln(z-z_i)$ at $z$ with respect to the
+    $z_i$ to write the potential at $z$ as
+    $$U(z) = \sum_i q_i\ln(z-z_i) = a_0 \ln z - \sum_{k=1}^M \frac{a_k}{z^k} + {\mathcal O} \left( \frac{R}{z} \right)^{M+1}$$
+    where
+    $$a_0 = \sum_i q_i {\rm ~~ and ~~} a_k = \sum_{i=1}^{N_c} \frac{q_i z_i^k}{k}, k\geq 1.$$
+    (Note that, for the potential, the real part of the right hand side
+    is to be taken.)
+
+    This is called a *multipole expansion*. A similar expansion exist in
+    three dimensions.
+
+10. [\[]{}D2[\]]{} In this problem, we consider the function $1/z$ close
+    to the real axis: $z=x-\rm i \epsilon$ where $\epsilon$ is small.
+    Show that the imaginary part of this function approaches $\pi$ times
+    the Dirac delta-function $\delta(x)$ for $\epsilon\rightarrow 0$. Do
+    this by showing that the integral over that function is $\pi$ and
+    that, when multiplied by a function $f(x)$, the result only depends
+    on $f(0)$.