complex analysis
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prerequisites:
table of contents:
intro to \(\mathbb{C}\)
complex numbers can be thought of as points in the coordinate plane, the same way that real numbers are points on the real line.so then, any complex number \(z\) can be thought of as an ordered pair of real numbers. $$z=(x,y),\qquad x,y\in\mathbb{R}$$ \(x\) is called the real part of \(z\), denoted \(\text{Re}(z)=x\).
\(y\) is called the imaginary part of \(z\), denoted \(\text{Im}(z)=y\).
thus, numbers of the form \((x,0)\) are on the real axis (and are real numbers), and those of the form \((0,y)\) (with \(y\neq0\)) are on the imaginary axis (and are called imaginary numbers).
together, the real and imaginary axes make up the complex plane, denoted \(\mathbb{C}\).
two complex numbers are equal 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 sum 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 product of two complex numbers is defined as follows: $$z_1 z_2=(x_1,y_1)(x_2,y_2)=(x_1 x_2-y_1 y_2,y_1 x_2+x_1 y_2)$$ you can easily remember complex multiplication as "\(\text{(lefts - rights , in + out)}\)."
these definitions of sum and product extended into \(\mathbb{C}\) become the usual sum and product when restricted back to \(\mathbb{R}\).
we can now deduce the following about any \(z\): $$ z=(x,y)=(x,0)+(0,y)\\ (0,1)(y,0)=(0y-1(0),1y-0(0))=(0,y)\\ \therefore z=(x,0)+(0,1)(y,0) $$ \((0,1)\) is called the imaginary unit, denoted \(i\).
this gives rise to the standard cartesian representation 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 results from the definition of multiplication in \(\mathbb{C}\)!
however, notice that the result is really \(i^2=-1\), so actually: $$ i^2=-1\\ \Rightarrow i=\sqrt{-1}\text{ and } i=-\sqrt{-1}\\ \Rightarrow i=\sqrt{-1}\text{ and} -i=\sqrt{-1}\\ \therefore i=-i \text{ ???} $$ this means that \(i\) and \(-i\) are algebraically indistinguishable, since choosing \(i=\sqrt{-1}\) or \(i=-\sqrt{-1}\) only affects the representations of complex numbers.
this is why we defined \(i\) with the coordinate representation \((0,1)\); it's totally unambiguous!
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 \(i^2=-1\).
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!
Like \(\mathbb{R}\), \(\mathbb{C}\) is a field:
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closure
- \(z_1+z_2\in\mathbb{C}\)
- \(z_1 z_2\in\mathbb{C}\)
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associativity
- \((z_1+z_2)+z_3=z_1+(z_2+z_3)\)
- \((z_1 z_2) z_3=z_1 (z_2 z_3)\)
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commutativity
- \(z_1+z_2=z_2+z_1\)
- \(z_1 z_2=z_2 z_1\)
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distributivity
- \(z(z_1+z_2)=z z_1+z z_2\)
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identities
- \(z+0=z\)
- \(z(1)=z\)
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inverses
- \(\forall z,\exists (-z)\) such that \(z+(-z)=0\)
- \(\forall z\neq0,\exists z^{-1}\) such that \(zz^{-1}=1\)
computing \(-z\): $$ z+(-z)=0\\ \text{Let }z=x+iy,(-z)=u+iv\\ \Rightarrow (x+iy)+(u+iv)=0\\ \Rightarrow (x+u)+i(y+v)=0+0i\\ \Rightarrow x+u=0,\qquad y+v=0\\ \Rightarrow -x=u,\qquad -y=v\\ \Rightarrow (-z)=(-x)+i(-y)=-x-iy=-(x+iy)\\ \therefore -z=-(z) $$ so, \(-z\) is defined uniquely for any \(z\).
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.
the existence of \(z^{-1}\) also implies that if \(z_1 z_2=0\), then \(z_1=0\) or \(z_2=0\). $$ z_1 z_2=0\\ \text{WLOG, let } z_1\neq0\\ \Rightarrow {z_1}^{-1} \text{ is defined}\\ \Rightarrow z_2=z_2(z_1 {z_1}^{-1})=(z_2 z_1) {z_1}^{-1}=(z_1 z_2) {z_1}^{-1}=0({z_1}^{-1})\\ \therefore z_2=0\\ $$ the difference of two complex numbers is defined in terms of additive inverses: $$z_1-z_2=z_1+(-z_2)$$ the quotient 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: $$ \frac{1}{z_1 z_2}=\left(\frac{1}{z_1}\right) \left(\frac{1}{z_2}\right) \text{ since } (z_1 z_2)(z_1 z_2)^{-1}=1=(z_1 z_1^{-1})(z_2 z_2^{-1})=(z_1 z_2)(z_1^{-1} z_2^{-1})\\ \\ \frac{z_1+z_2}{z_3}=\frac{z_1}{z_3}+\frac{z_2}{z_3} \text{ since } (z_1+z_2){z_3}^{-1}=z_1{z_3}^{-1}+z_2{z_3}^{-1} $$ the cancellation law (\(\frac{z_1 z}{z_2 z}=\frac{z_1}{z_2}\)) and the double inverse law (\(\frac{1}{1/z}=(z^{-1})^{-1}=1\)) also follow.
the binomial formula also extends into \(\mathbb{C}\): $$ (z_1+z_2)^n=\sum_{k=0}^{n}\binom{n}{k} {z_1}^k {z_2}^{n-k},\qquad z_1\neq0,z_2\neq0,n=1,2,3,...\\ \text{where } \binom{n}{k}=\frac{n!}{k!(n-k)!},\qquad k=0,1,2,...,n $$ as well as the triangle inequalities: $$ ||z_1|-|z_2||\leq|z_1 + z_2|\leq|z_1|+|z_2| $$ since complex numbers can be represented as points on the complex plane, certain geometric intuitions can be applied to them.
for example, adding two complex numbers is the same as adding two vectors together, tip-to-tail (i.e. coordinate-wise).
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 modulus 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 \(\mathbb{C}\)!
naturally, modulus becomes absolute value when restricted back to \(\mathbb{R}\).
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.
if you reflect a complex number \(z\) over the real axis, you reach the conjugate of \(z\), denoted \(\overline{z}\). $$ z=x+iy\Rightarrow \overline{z}=x-iy $$ conjugation holds many important properties:
- double conjugation law: \(\overline{(\overline{z})}=z\)
- same modulus as \(z\): \(|\overline{z}|=|z|\)
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distributes over the basic operations:
- \(\overline{z_1+z_2}=\overline{z_1}+\overline{z_2}\)
- \(\overline{z_1-z_2}=\overline{z_1}-\overline{z_2}\)
- \(\overline{z_1 z_2}=\overline{z_1}\cdot \overline{z_2}\)
- \(\overline{\left(\frac{z_1}{z_2}\right)}=\frac{\overline{z_1}}{\overline{z_2}}\)
- \(\text{Re}(z)=\frac{z+\overline{z}}{2}\) and \(\text{Im}(z)=\frac{z-\overline{z}}{2i}\)
- \(z \overline{z}=|z|^2=x^2+y^2\)
this last property is very useful!
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 always results in a real number!
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