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        <h1>complex analysis</h1>
        
        <a href="./index.html">go home</a></li>

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            prerequisites:
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                <li><a href="./calculus.html">🟥calculus</a></li>
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            table of contents:
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                <li><a href="#intro-to-C">🟨intro to \(\mathbb{C}\)</a></li>
                <li><a href="#functions-of-z">🟥functions of \(z\)</a></li>
                <li><a href="#derivatives">🟥derivatives</a></li>
                <li><a href="#elementary-functions">🟥elementary functions</a></li>
                <li><a href="#integrals">🟥integrals</a></li>
                <li><a href="#series">🟥series</a></li>
                <li><a href="#singularities-and-residues">🟥singularities and residues</a></li>
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        you may remember the notion of "complex numbers" \(a+bi\), where \(i:=\sqrt{-1}\),
        since they were briefly mentioned in <a href="./algebra.html">🟥elementary algebra</a>
        (for example, as a way to find the roots of the polynomial \(x^2+1=0\)).
            <br><br>
        this page will formally reintroduce complex numbers and teach you how to think about them geometrically and do calculus with them!

        <hr>

        <h2 id="intro-to-C">intro to \(\mathbb{C}\)</h2>

        complex numbers can be thought of as points in the coordinate plane, the same way that real numbers are points on the real line.
            <br><br>
        so, any <b>complex number</b> \(z\) can be thought of as an ordered pair of real numbers.
            $$z=(x,y),\qquad x,y\in\mathbb{R}$$
        \(x\) is called the <b>real part</b> of \(z\), denoted \(\text{Re}(z)=x\).
            <br>
        \(y\) is called the <b>imaginary part</b> of \(z\), denoted \(\text{Im}(z)=y\).
            <br><br>
        thus, numbers of the form \((x,0)\) are on the <b>real axis</b> (and are <b>real numbers</b>), and those of the form \((0,y)\) (with \(y\neq0\))
        are on the <b>imaginary axis</b> (and are called <b>imaginary numbers</b>).
            <br><br>
        together, the real and imaginary axes make up the <b>complex plane</b>; 
        it represents the set of all complex numbers, denoted \(\mathbb{C}\).
            <br><br>
        two complex numbers are <b>equal</b> when their real parts are equal and their imaginary parts are equal
        (i.e., they are the same point on the complex plane).
            $$
                z_1=z_2\Leftrightarrow
                \begin{cases}
                    {x_1=x_2\\
                    y_1=y_2}
                \end{cases}
            $$
        the <b>sum</b> of two complex numbers is the coordinate-wise addition of their parts:
            $$z_1+z_2=(x_1,y_1)+(x_2,y_2)=(x_1+x_2,y_1+y_2)$$
        the <b>product</b> of two complex numbers is defined as follows:
            $$z_1 z_2=(x_1,y_1)(x_2,y_2)=(x_1 x_2-y_1 y_2,y_1 x_2+x_1 y_2)$$
        i like to remember complex multiplication as "\(\text{(lefts - rights , in + out)}\)"!
            <br><br>
        these definitions of sum and product extended into \(\mathbb{C}\) become the "usual" (real-valued) sum and product when restricted back to \(\mathbb{R}\).
            <br><br>
        from the product, we have exponentiation of a complex number defined in terms of repeated multiplication:
            $$z^n=(x,y)^n=(x,y)(x,y)(x,y)...(x,y) \text{ (}n\text{ times)},\qquad n=1,2,3,...$$
        the \((x,y)\) <b>coordinate-pair form</b> of complex numbers works fine,
        but there is a more useful representation of any complex number \(z\) that we can derive using the sum and product we just defined:
            $$
                z=(x,y)=(x,0)+(0,y)\\
                (0,1)(y,0)=(0y-1(0),1y-0(0))=(0,y)\\\\
                \therefore z=(x,0)+(0,1)(y,0)
            $$
        \((0,1)\) is called the <b>imaginary unit</b>, denoted \(i\).
            <br><br>
        this gives rise to the standard <b>cartesian form</b> of a complex number \(z\):
            $$z=x+iy,\qquad x,y\in\mathbb{R}$$
        computing \(i^2\):
            $$
                i^2=i\cdot i=(0,1)(0,1)=(0(0)-1(1),0(1)+1(0))=(-1,0)=-1\\\\
                \therefore i^2=-1
            $$
        so, \(i=\sqrt{-1}\) is a property that <i>results</i> from the definition of multiplication in \(\mathbb{C}\)!
            <br>
        however, notice that the result is really \(i^2=-1\), so <i>actually</i>:
            $$
                i^2=-1\\
                \Rightarrow i=\sqrt{-1}\text{ and } i=-\sqrt{-1}\\
                \Rightarrow i=\sqrt{-1}\text{ and} -i=\sqrt{-1}\\\\
                \therefore i=-i \text{ ???}
            $$
        this means that \(i\) and \(-i\) are <b>algebraically indistinguishable</b>, since choosing \(i=\sqrt{-1}\) or \(i=-\sqrt{-1}\)
        only affects the <i>representations</i> of complex numbers.
            <br><br>
        <i>this</i> is why we defined \(i\) with the coordinate-pair representation \((0,1)\); it's totally unambiguous!
            <br><br>
        since all complex numbers are made of real numbers with a potential added factor of \(i\), you can basically treat \(i\) as a variable with the added condition that \(i^2=-1\).
            <br><br>
        so, the sum of complex numbers can be treated in this way:
            $$
                z_1+z_2=(x_1+iy_1)+(x_2+iy_2)\\
                =(x_1+x_2)+(iy_1+iy_2)\\\\
                \therefore z_1+z_2=(x_1+x_2)+i(y_1+y_2)
            $$
        and the product of complex numbers can be treated in this way (FOIL):
            $$
                z_1 z_2=(x_1+iy_1)(x_2+iy_2)\\
                =x_1 x_2+iy_1(x_2)+x_1(iy_2)+(i y_1)(i y_2)\\
                =x_1 x_2+i(y_1 x_2)+i(x_1 y_2)+i^2 (y_1 y_2)\\
                =x_1 x_2+i(y_1 x_2 + x_1 y_2)- y_1 y_2\\\\
                \therefore z_1 z_2=(x_1 x_2 - y_1 y_2)+i(y_1x_2+x_1y_2)
            $$
        which is a lot more straightforward than remembering their formal definitions!
            <br><br>
        Like \(\mathbb{R}\), \(\mathbb{C}\) is a field, so  it also has the following properties:
        <ol>
            <li>
                closure
                <ul>
                    <li>\(z_1+z_2\in\mathbb{C}\)</li>
                    <li>\(z_1 z_2\in\mathbb{C}\)</li>
                </ul>
            </li>
            <li>
                associativity
                <ul>
                    <li>\((z_1+z_2)+z_3=z_1+(z_2+z_3)\)</li>
                    <li>\((z_1 z_2) z_3=z_1 (z_2 z_3)\)</li>
                </ul>
            </li>
            <li>
                commutativity
                <ul>
                    <li>\(z_1+z_2=z_2+z_1\)</li>
                    <li>\(z_1 z_2=z_2 z_1\)</li>
                </ul>
            </li>
            <li>
                distributivity
                <ul>
                    <li>\(z(z_1+z_2)=z z_1+z z_2\)</li>
                </ul>
            </li>
            <li>
                identities
                <ul>
                    <li>\(z+0=z\)</li>
                    <li>\(z(1)=z\)</li>
                </ul>
            </li>
            <li>
                inverses
                <ul>
                    <li>\(\forall z,\exists (-z)\) such that \(z+(-z)=0\)</li>
                    <li>\(\forall z\neq0,\exists z^{-1}\) such that \(zz^{-1}=1\)</li>
                </ul>
            </li>
        </ol>
            <br>
        computing \(-z\):
            $$
                z+(-z)=0\\
                \text{Let }z=x+iy,(-z)=u+iv\\
                \Rightarrow (x+iy)+(u+iv)=0\\
                \Rightarrow (x+u)+i(y+v)=0+0i\\
                \Rightarrow x+u=0,\qquad y+v=0\\
                \Rightarrow -x=u,\qquad -y=v\\
                \Rightarrow (-z)=(-x)+i(-y)=-x-iy=-(x+iy)\\\\
                \therefore -z=-(z)
            $$
        who would have guessed?
            <br><br>
        we could also compute \(z^{-1}\) right now, but we'll hold off on that until we have some more useful definitions to make the computation easier.
            <br><br>
        the existence of \(z^{-1}\) also implies that the <b>zero-product property</b> extends into \(\mathbb{C}\) (\(z_1 z_2=0\Rightarrow z_1=0\vee z_2=0\)).
            $$
                z_1 z_2=0\\
                \text{WLOG, let } z_1\neq0\\
                \Rightarrow {z_1}^{-1} \text{ is defined}\\
                \Rightarrow z_2=z_2(z_1 {z_1}^{-1})=(z_2 z_1) {z_1}^{-1}=(z_1 z_2) {z_1}^{-1}=0({z_1}^{-1})\\\\
                \therefore z_2=0\\
            $$
        the <b>difference</b> of two complex numbers is defined in terms of additive inverses:
            $$z_1-z_2=z_1+(-z_2)$$
        and the <b>quotient</b> of two complex numbers is defined in terms of multiplicative inverses (where \(\frac{1}{z}:={z}^{-1}\)):
            $$\frac{z_1}{z_2}=z_1 \left(\frac{1}{z_2}\right)=z_1 {z_2}^{-1}$$
        all the usual properties of quotients in \(\mathbb{R}\) extend into \(\mathbb{C}\), such as:
        <ul>
            <li>\(\frac{1}{z_1 z_2}=\left(\frac{1}{z_1}\right) \left(\frac{1}{z_2}\right)\)</li>
            <li>\(\frac{z_1+z_2}{z_3}=\frac{z_1}{z_3}+\frac{z_2}{z_3}\)</li>
            <li><b>cancellation law</b>: \(\frac{z_1 z}{z_2 z}=\frac{z_1}{z_2}\)</li>
            <li><b>double inverse law</b>: \(\frac{1}{1/z}=(z^{-1})^{-1}=1\)</li>
        </ul>
            <br>
        the <b>binomial formula</b> also extends into \(\mathbb{C}\):
            $$
                (z_1+z_2)^n=\sum_{k=0}^{n}\binom{n}{k} {z_1}^k {z_2}^{n-k},\qquad z_1\neq0,z_2\neq0,n=1,2,3,...
            $$
        as well as the <b>triangle inequalities</b>:
            $$
                ||z_1|-|z_2||\leq|z_1 + z_2|\leq|z_1|+|z_2|
            $$
        since complex numbers can be represented as points on the complex plane, certain geometric intuitions you are already familar with can be applied to them.
            <br><br>
        for example, adding two complex numbers is the same as adding two vectors together tip-to-tail (i.e. coordinate-wise).
            <br><br>
        if you compute the distance from a point \(z\) on the complex plane to \((0,0)\), you will get a non-negative real number called the <b>modulus</b> of \(z\), denoted \(|z|\).
            $$
                |z|=|x+iy|=\sqrt{x^2+y^2}
            $$
        if you recognize the notation, that is beacuse the modulus extends the concept of absolute value in \(\mathbb{R}\) to \(\mathbb{C}\)!
            <br>
        naturally, modulus becomes absolute value when restricted back to \(\mathbb{R}\).
            <br><br>
        also, these inequalities follow directly from the definition of the modulus:
            $$
                \text{Re}(z)\leq|\text{Re}(z)|\leq|z|\\
                \text{Im}(z)\leq|\text{Im}(z)|\leq|z|
            $$
        i.e., the real/imaginary parts of \(z\) will never exceed the total length of \(z\), the same way each leg of a right triangle will never be longer than its hypotenuse.
            <br><br>
        if you reflect a complex number \(z\) over the real axis, you reach the <b>conjugate</b> of \(z\), denoted \(\overline{z}\).
            $$
                z=x+iy\quad\Leftrightarrow\quad\overline{z}=x-iy
            $$
        conjugation holds many important properties:
        <ul>
            <li><b>double conjugation law</b>: \(\overline{(\overline{z})}=z\)</li>
            <li>same modulus as \(z\): \(|\overline{z}|=|z|\)</li>
            <li>
                distributes over the basic operations:
                <ul>
                    <li>\(\overline{z_1+z_2}=\overline{z_1}+\overline{z_2}\)</li>
                    <li>\(\overline{z_1-z_2}=\overline{z_1}-\overline{z_2}\)</li>
                    <li>\(\overline{z_1 z_2}=\overline{z_1}\cdot \overline{z_2}\)</li>
                    <li>\(\overline{\left(\frac{z_1}{z_2}\right)}=\frac{\overline{z_1}}{\overline{z_2}}\)</li>
                </ul>
            </li>
            <li>\(\text{Re}(z)=\frac{z+\overline{z}}{2}\) and \(\text{Im}(z)=\frac{z-\overline{z}}{2i}\)</li>
            <li>\(z \overline{z}=|z|^2=x^2+y^2\)</li>
        </ul>
            <br>
        this last property is <i>very</i> useful!
            <br>
        you can use it to, for example, show that the modulus is multiplicative:
            $$
                |z_1 z_2|^2=(z_1 z_2)\overline{(z_1 z_2)}\\
                =(z_1 z_2)(\overline{z_1}\cdot \overline{z_2})\\
                =(z_1\overline{z_1})(z_2 \overline{z_2})\\
                =|z_1|^2 |z_2|^2\\
                \therefore |z_1 z_2|=|z_1||z_2|
            $$
        and also to easily compute \(z^{-1}\):
            $$
                z^{-1}=\frac{1}{z}=\frac{1}{z} \frac{\overline{z}}{\overline{z}}=\frac{\overline{z}}{z \overline{z}}=\frac{\overline{z}}{|z|^2}\\\\
                \therefore z^{-1}=\frac{\overline{z}}{|z|^2}=\frac{x}{x^2+y^2}-i \frac{y}{x^2+y^2}
            $$
        overall, if you are ever dealing with computations where some non-real number is in a place you don't want it to be (such as in the denominator of a fraction),
        multiplying by the conjugate is a very useful technique since multiplying a complex number by its conjugate <i>always</i> results in a real number!
            <br><br>
        we've spent a lot of time on the cartesian form of complex numbers, but same as the standard \(xy\)-coordinate plane,
        we can also think of complex numbers in a <i>polar</i> coordinate system.
            <br><br>
        that is, instead of thinking of \(z\) in terms of \(x\) and \(y\) coordinates,
        we can consider \(z\) in terms of the distance from the origin \(r\) and its angle from the positive real axis \(\theta\).
            $$
                \text{Re}(z)=x=r\cos{(\theta)},\qquad \text{Im}(z)=y=r\sin{(\theta)}
            $$
        so, the <b>polar form</b> of any complex number \(z\) is:
            $$
                z=r(\cos{(\theta)}+i\sin{(\theta)}),\qquad r=|z|,\theta=\tan{\left(\frac{y}{x}\right)}
            $$
        note a couple of things:
        <ul>
            <li>
                \(\tan{\left(\frac{y}{x}\right)}=\tan{\left(\frac{-y}{-x}\right)}\), so we need to keep track of which quadrant \(z\) is in by looking at \(x\) and \(y\).
                <ul><li>you better have the unit circle memorized!</li></ul>
            </li><br>
            <li>
                \(\theta=\tan{\left(\frac{y}{x}\right)}\) has infinitely many solutions.
                <ul><li>if you rotate any complex number by \(2\pi\), it'll go back to the same spot it was originally in, except its angle will be \(2\pi\) greater; \(\theta\) is \(2\pi\)-periodic!</li></ul>
            </li>
        </ul>
            <br>
        any one of these infinite solutions for \(\theta\) is called an <b>argument</b> of \(z\), and the set of all arguments of \(z\) is denoted \(\text{arg}(z)\).
            <br><br>
        if you restrict the value of \(\theta\) to \((-\pi,\pi]\), you get the <b>principal argument</b> of \(z\), denoted \(\text{Arg}(z)\).
        <ul><li>\(-\pi\) isn't included to avoid ambiguity on what the principal argument of a negative real number is: it'll always be \(\pi\).</li></ul>
            <br>
        as a shorthand for polar form, let's introduce this notation, called the <b>Euler symbol</b>:
            $$e^{i\theta}=\cos{(\theta)}+i\sin{(\theta)}$$
        note that \(e^{i\theta}\) only denotes <b>unit numbers</b> (numbers with modulus \(1\)):
            $$
                |e^{i\theta}|=|\cos{(\theta)}+i\sin{(\theta)}|
                =\sqrt{\cos^2{(\theta)}+\sin^2{(\theta)}}
                =\sqrt{1}
                =1
            $$
        so, any complex number \(z\) can be written like this (called its <b>exponential form</b>):
            $$z=re^{i\theta}$$
        surprisingly, a complex number's <i>exponential</i> form seems to have an <i>exponential</i> function in it (<i>foreshadowing</i>),
        but for now, think of it simply as notation.
            <br><br>
        computing \(e^{i\theta}\) for some friendly values of \(\theta\):
            $$
                e^{i(0)}=\cos{(0)}+i\sin{(0)}=1+0i=1\\
                e^{i\left(\frac{\pi}{2}\right)}=\cos{\left(\frac{\pi}{2}\right)}+i\sin{\left(\frac{\pi}{2}\right)}=0+1i=i\\
                e^{i(\pi)}=\cos{(\pi)}+i\sin{(\pi)}=-1+0i=-1\\
                e^{i\left(\frac{-\pi}{2}\right)}=\cos{\left(\frac{-\pi}{2}\right)}+i\sin{\left(\frac{-\pi}{2}\right)}=0+(-1)i=-i\\
            $$
        some properties of the Euler symbol that are useful:
        <ul>
            <li><b>exponential property</b>: \(e^{i\theta_1}e^{i\theta_2}=e^{i(\theta_1+\theta_2)}\)</li>
            <li>its conjugate is its inverse: \(\overline{(e^{i\theta})}=e^{-i\theta}=\frac{1}{e^{i\theta}}\)</li>
        </ul>
            <br>
        both the cartesian and exponential forms have their own advantages and disadvantages during computation,
        so it is important to know how to transition between them using the polar form:
            $$
                \begin{array}{c|c|c}
                    \text{cartesian} & \text{polar} & \text{exponential}\\
                    \hline z=x+iy & \begin{matrix}z=\sqrt{x^2+y^2}(\frac{x}{\sqrt{x^2+y^2}}+i\frac{y}{\sqrt{x^2+y^2}})\\=r(\cos{(\theta)}+i\sin{(\theta)})\end{matrix} & z=re^{i\theta}
                \end{array}
            $$
        now that we've defined complex numbers, basic operations using them, and their different representations, we should
        see how it all looks geometrically!
            <br><br>
        here's a graphic summarizing the representations of any complex number:
            $$\text{visual of coordinate-pair,cartesian,polar,exp}$$
        as mentioned before, addition and subtraction are the same as for vectors:
            $$
                \begin{matrix}
                    \text{pic } +\\
                    z_1+z_2=(x_1+x_2)+i(y_1+y_2)
                \end{matrix}
                \qquad
                \begin{matrix}
                    \text{pic } -\\
                    z_1-z_2=(x_1-x_2)+i(y_1-y_2)
                \end{matrix}
            $$
        since the Euler symbol has the exponential property, multiplication and division simply multiplies or divides moduli and adds or subtracts arguments:
            $$
                \begin{matrix}
                    \text{pic } \cdot\\
                    z_1 z_2=r_1 r_2 e^{i(\theta_1+\theta_2)}
                \end{matrix}
                \qquad
                \begin{matrix}
                    \text{pic }/\\
                    \frac{z_1}{z_2}=\frac{r_1}{r_2} e^{i(\theta_1-\theta_2)}
                \end{matrix}
            $$
        and finally, as mentioned before, conjugation reflects across the real axis:
            $$\text{reflection across Re... do i mention that "negatives" reflect across the line y=-x now?}$$
        since we are thinking geometrically, we should also define some important regions of the complex plane and terminology relating to those regions.
            <br><br>
        for \(\epsilon>0\), the \(\epsilon\)<b>-neighborhood</b> of \(z_0\) (denoted \(B_{\epsilon}(z_0)\)) is the open disk of radius \(\epsilon\) centered at \(z_0\).
            $$B_{\epsilon}(z_0)=\{z\in\mathbb{C}:|z-z_0|<\epsilon\}$$
        the <b>deleted</b> (or <b>punctured</b>) \(\epsilon\)-neighborhood of \(z_0\) is the same as above, except it doesn't include \(z_0\) itself.
            $$B^*_{\epsilon}(z_0)=B_{\epsilon}(z_0)\backslash\{z_0\}$$
        these are often shortened simply "neighborhoods" and "deleted neighborhoods" when their radii are arbitrary.
            <br><br>
        for any \(S\subseteq\mathbb{C}\):
        <ul>
            <li>\(z_0\) is an <b>interior</b> point of \(S\) if there is a neighborhood around it that lies entirely <i>inside</i> of \(S\).</li>
            <li>\(z_0\) is an <b>exterior</b> point of \(S\) if there is a neighborhood around it that lies entirely <i>outside</i> of \(S\).</li>
            <li>\(z_0\) is a <b>boundary</b> point of \(S\) if it's niether interior nor exterior to \(S\) (i.e. every nieghborhood around it contains interior and exterior points of \(S\)).</li>
        </ul>
            $$\text{the amazing image i have in my notes}$$
        naturally, the <b>interior</b> of \(S\) is the set of all its interior points,
        the <b>exterior</b> of \(S\) is the set of all its exterior points,
        and the <b>boundary</b> of \(S\) is the set of all its boundary points (denoted \(\partial S\)).
        <ul>
            <li>
                \(S\) is <b>open</b> if every point in \(S\) is an interior point.
                <ul><li>i.e., no part of \(\partial S\) is in \(S\).</li></ul>
            </li>
            <li>
                \(S\) is <b>closed</b> if its complement \(\mathbb{C}\backslash S\) is open.
                <ul><li>i.e., all of \(\partial S\) is in \(S\)</li><li>\(\partial S\cup S\) is called the <b>closure</b> of \(S\).</li></ul>
            </li>
        </ul>
            <br>
        <i>sets are not doors</i>: they can be both closed and open (creatively denoted "<b>clopen</b>"),
        or niether open nor closed (there's no standard name for this, but i'll call a set like this a <b>NOOC</b>, short for "not open or closed").
            $$\text{pic of open,closed sets and closure of a set}$$
        <ul>
            <li>\(S\) is <b>convex</b> if any two points in it can be connected by a single straight line that lies entirely in it.</li>
            <li>\(S\) is <b>connected</b> if any two points in it can be connected by a finite number of straight lines which share endpoints (called a <b>polygonal path</b>) that lie entirely in it.</li>
        </ul>
            $$\text{another W pic}$$
        <ul>
            <li>\(S\) is <b>bounded</b> if there exists a neighborhood centered at \(0\) that contains all of \(S\).</li>
        </ul>
            $$\text{another W pic}$$
        <ul>
            <li>\(S\) is a <b>domain</b> if it is open and connected.</li>
            <li>
                \(S\) is a <b>region</b> if it is connected and all its points are either interior or boundary points.
                <ul><li>i.e., it's a domain, plus possibly some part of its boundary.</li></ul>
            </li>
        </ul>
            $$\text{another W pic}$$
        finally, the <b>accumulation set</b> of \(S\) (denoted \(\text{Acc}(S)\)) is the set of all points \(z\) where an arbitrarily
        small deleted neighborhood around \(z\) intersects \(S\).
            $$\text{Acc}(S)=\{z\in\mathbb{C}:\forall\epsilon>0,B^*_\epsilon(z)\cap S\neq\emptyset\}$$
        think of \(\text{Acc}(S)\) as the set of points which \(S\) gets arbitrarily close to, and which may or may not be in S.
            $$\text{maybe a W pic?}$$
        any point in \(\text{Acc}(S)\) is called an <b>accumulation point</b> or a <b>limit point</b>.
        <hr>

        <h2 id="functions-of-z">functions of \(z\)</h2>
        
        <span style="color: var(--woofleBlue);">
            funcs of z, lims, cont
        </span>

        <hr> 

        <h2 id="derivatives">derivatives</h2>
        
        <span style="color: var(--woofleBlue);">
            derivs, CR eqs (cartesian, polar, complex), "analytic" and its consequences
        </span>

        <hr>

        <h2 id="elementary-functions">elementary functions</h2>
        
        <span style="color: var(--woofleBlue);">
            explain roots, powers, and ln here probably
        </span>
        
        <hr>

        <h2 id="integrals">integrals</h2>

        <span style="color: var(--woofleBlue);">
            the wonderful world of integrals and contour integrals 
        </span>

        <hr>

        <h2 id="series">series</h2>

        <span style="color: var(--woofleBlue);">
            power series, taylor and laurent series
        </span>

        <hr>

        <h2 id="singularities-and-residues">singularities and residues</h2>

        <span style="color: var(--woofleBlue);">
            cauchy residue thm and all that stuff that goes with it
        </span>

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