W15 - Notes

Complex algebra

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• Complex numbers basics

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01 Theory - Complex arithmetic

Theory 1

The complex numbers $ℂ$ are sums of real and imaginary numbers. Every complex number can be written uniquely in ‘Cartesian’ form:

$$z=a+bi,a,b\in ℝ$$

To add, subtract, scale, and multiply complex numbers, treat ‘$i$’ like a constant.

Simplify the result using $i^2=-1$.

For example:

$$\begin{array}{c}(1+3i)(2-2i)\gg \gg 2-2i+6i-6i^2\\ \\ \gg \gg 2+4i-6(-1)\gg \gg 8+4i\end{array}$$

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Complex conjugate

Every complex number has a complex conjugate:

$$z=a+bi\gg \gg \bar{z}=a-bi$$

For example:

$$\begin{array}{cc}\overline{2+5i}& =2-5i\\ & \\ \overline{2-5i}& =2+5i\end{array}$$

In general, $\overline{\stackrel{‾}{z}}=z$.

Conjugates are useful mainly because they eliminate imaginary parts:

$$(2+5i)(2-5i)\gg \gg 4+25\gg \gg 29$$

In general:

$$(a+bi)(a-bi)\gg \gg a^2-abi+bia-b^2{i}^2\gg \gg a^2+b^2\in ℝ$$

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Complex division

To divide complex numbers, use the conjugate to eliminate the imaginary part in the denominator.

For example, reciprocals:

$$\begin{array}{c}\frac{1}{a+bi}\gg \gg \frac{1}{a+bi}\cdot \frac{a-bi}{a-bi}\\ \\ \gg \gg \frac{a-bi}{a^2+b^2}\gg \gg \left(\frac{a}{a^2+b^2}\right)+\left(\frac{-b}{a^2+b^2}\right)i\end{array}$$

More general fractions:

$$\begin{array}{c}\frac{a+bi}{c+di}\gg \gg \frac{a+bi}{c+di}\cdot \frac{c-di}{c-di}\\ \\ \gg \gg \frac{ac+bd+(bc-ad)i}{c^2+d^2}\gg \gg \frac{ac+bd}{c^2+d^2}+\frac{bc-ad}{c^2+d^2}i\end{array}$$

Multiplication preserves conjugation

For any $z,w\in ℂ$:

$$\overline{zw}=\bar{z}\bar{w}$$

Therefore, one can take products or conjugates in either order.

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02 Illustration

Example - Complex multiplication

Complex multiplication

Compute the products:

(a) $(1-i)(4-7i)$ $$ (b) $(2+5i)(2-5i)$

Solution

(a) $(1-i)(4-7i)$

Expand:

$$(1-i)(4-7i)\gg \gg 4-7i-4i+7i^2$$

Simplify $i^2$:

$$\gg \gg 4-7i-4i+7(-1)\gg \gg -3-11i$$

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(b) $(2+5i)(2-5i)$

Expand:

$$(2+5i)(2-5i)\gg \gg 4-10i+10i-25i^2$$

Simplify $i^2$:

$$\gg \gg 4-10i+10i-25(-1)\gg \gg 29$$ Link to original

Example - Complex division

Complex division

Compute the following divisions of complex numbers:

(a) ${\displaystyle \frac{1}{-3+i}}$ $$ (b) ${\displaystyle \frac{1}{i}}$ $$ (c) ${\displaystyle \frac{1}{7i}}$ $$ (d) ${\displaystyle \frac{2+5i}{2-5i}}$

Solution

(a) ${\displaystyle \frac{1}{-3+i}}$

Conjugate is $-3-i$:

$$\frac{1}{-3+i}\gg \gg \frac{1}{-3+i}\cdot \frac{-3-i}{-3-i}$$

Simplify:

$$\gg \gg \frac{-3-i}{9+1}\gg \gg \frac{-3}{10}+\frac{-1}{10}i$$

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(b) ${\displaystyle \frac{1}{i}}$

Conjugate is $-i$:

$$\frac{1}{i}\gg \gg \frac{1}{i}\cdot \frac{-i}{-i}\gg \gg -i$$

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(c) ${\displaystyle \frac{1}{7i}}$

Factor out the $1/7$:

$$\frac{1}{7i}\gg \gg \frac{1}{7}\cdot \frac{1}{i}$$

Use $\frac{1}{i}=-i$:

$$\gg \gg \frac{1}{7}\cdot (-i)\gg \gg \frac{-1}{7}i$$

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(d) ${\displaystyle \frac{2+5i}{2-5i}}$

Denominator conjugate is $2+5i$:

$$\frac{2+5i}{2-5i}\gg \gg \frac{2+5i}{2-5i}\cdot \frac{2+5i}{2+5i}$$

Simplify:

$$\gg \gg \frac{4+20i+25i^2}{4+25}\gg \gg \frac{-21}{29}+\frac{20}{29}i$$Link to original

Complex exponential

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03 Theory - Complex plane and polar data

Theory 1

A complex number $z=x+iy$ can be represented in the plane as the point with Cartesian coordinates $(x,y)$. The coefficient of “$i$” determines the vertical coordinate, and the coefficient of “$1$” determines the horizontal coordinate.

Let us be given a complex number $z=a+bi$.

The “real part” and “imaginary part” of $z$ can be extracted with designated functions:

$$\begin{array}{c}\mathrm{Re}(z)=a,\mathrm{Im}(z)=b,\text{for }z=a+bi\\ \\ z=\mathrm{Re}(z)+\mathrm{Im}(z)i\end{array}$$

The polar data (radius and angle) have special names and notations for complex numbers:

$$\begin{array}{cc}r& =\sqrt{a^2+b^2}\\ & \\ & =|z|\\ & \\ & =\text{“modulus” of }z\end{array}\begin{array}{cc}\theta & ={\tan }^{-1}(b/a)\stackrel{?}{+}\pi \\ & \\ & =\mathrm{Arg}(z)\\ & \\ & =\text{“argument” of }z\end{array}$$

Using this notation, we see that product with the conjugate gives square of modulus:

$$z\bar{z}=|z{|}^2$$Link to original

04 Theory - cis, Euler, products, powers

Theory 1

Multiplication of complex numbers is much easier to understand when the numbers are written using polar form.

There is a shorthand ‘$\mathrm{cis}$’ notation. Convert to polar coordinates, so $a=r\cos \theta$ and $b=r\sin \theta$:

$$\begin{array}{cc}a+bi& \gg \gg r\cos \theta +r\sin \theta i\\ & \\ & \gg \gg r(\cos \theta +i\sin \theta )\\ & \\ & \gg \gg r\mathrm{cis}\theta \end{array}$$

The “$\mathrm{cis}$” stands for $\cos \theta +i\sin \theta$. For example:

$$\begin{array}{c}\sqrt{2}-\sqrt{2}i\gg \gg 2(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i)\\ \\ \gg \gg 2\cos (-\frac{\pi }{4})+2\sin (-\frac{\pi }{4})i\\ \\ \gg \gg 2\mathrm{cis}(-\frac{\pi }{4})\end{array}$$

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Euler Formula

General Euler Formula:

$$r{e}^{i\theta }=r\cos \theta +ir\sin \theta$$

On the unit circle $r=1$:

$$e^{i\theta }=\cos \theta +i\sin \theta$$

The form $r{e}^{i\theta }$ expresses the same data as the $\mathrm{cis}$ form. The principal advantage of the form $r{e}^{i\theta }$ is that it reveals the rule for multiplication, which comes from exponent laws:

Complex multiplication - Exponential form

$$r_1{e}^{i{\theta }_1}\cdot r_2{e}^{i{\theta }_2}=(r_1{r}_2)e^{i({\theta }_1+{\theta }_2)}$$

In words:

• Multiply radii
• Add angles

Notice:

$$\text{multiply by }{e}^{i\frac{\pi }{2}}⟺\text{rotate by }+{90}^{\circ }$$

Notice:

$$e^{i\frac{\pi }{2}}=+i$$

Therefore $i$ ‘acts upon’ other numbers by rotating them ${90}^{\circ }$ counterclockwise!

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De Moivre’s Theorem - Complex powers

In exponential notation:

$$(r{e}^{i\theta }{)}^n=r^n{e}^{i\cdot n\theta }$$

In $\mathrm{cis}$ notation:

$$(r\mathrm{cis}\theta {)}^n=r^n\mathrm{cis}(n\theta )$$

Expanded $\mathrm{cis}$ notation:

$$(r\cos \theta +ir\sin \theta {)}^n=r^n\cos (n\theta )+i{r}^n\sin (n\theta )$$

So the power of $n$ acts like this:

• Stretch: $r$ to $r^n$
• Rotate: by $n$ increments of $\theta$

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Extra - Derivation of Euler Formula

Recall the power series for $e^x$:

$$e^x=1+\frac{1}{1!}x+\frac{1}{2!}{x}^2+\frac{1}{3!}{x}^3+\dots =\sum _{i=0}^{\infty }\frac{1}{i!}{x}^i$$

Plug in $x=i\theta$:

$$\begin{array}{c}{e}^{i\theta }\gg \gg 1+(i\theta )+\frac{1}{2!}{(i\theta )}^2+\frac{1}{3!}{(i\theta )}^3+\cdots +\end{array}$$

Simplify terms:

$$\gg \gg 1+i\theta -\frac{1}{2!}{\theta }^2-\frac{1}{3!}i{\theta }^3+\frac{1}{4!}{\theta }^4+\frac{1}{5!}i{\theta }^5-\frac{1}{6!}{\theta }^6-\frac{1}{7!}i{\theta }^7+\frac{1}{8!}{\theta }^8+\cdots$$

Separate by $i$-factor. Select out the $\overbrace{\text{terms with }i}$:

$$\gg \gg 1\overbrace{+i\theta }-\frac{1}{2!}{\theta }^2\overbrace{-\frac{1}{3!}i{\theta }^3}+\frac{1}{4!}{\theta }^4\overbrace{+\frac{1}{5!}i{\theta }^5}-\frac{1}{6!}{\theta }^6\overbrace{-\frac{1}{7!}i{\theta }^7}+\frac{1}{8!}{\theta }^8+\cdots$$

Separate into a series without $i$ and a series with $i$:

$$\gg \gg (1-\frac{1}{2!}{\theta }^2+\frac{1}{4!}{\theta }^4-\cdots )+(\theta -\frac{1}{3!}{\theta }^3+\frac{1}{5!}{\theta }^5-\cdots )i$$

Identify $\cos \theta$ and $\sin \theta i$. Write trig series:

$$\begin{array}{cc}\cos \theta & =1-\frac{1}{2!}{\theta }^2+\frac{1}{4!}{\theta }^4-\cdots \\ & \\ \sin \theta & =\theta -\frac{1}{3!}{\theta }^3+\frac{1}{5!}{\theta }^5-\cdots \end{array}$$

Therefore $e^{i\theta }=\cos \theta +i\sin \theta$.

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05 Illustration

Example - Complex product, quotient, power using Euler

Complex product, quotient, power using Euler

Define:

$$z=2e^{i\frac{\pi }{2}}w=5e^{i\frac{\pi }{3}}$$

Product $zw$:

$$\begin{array}{c}zw\gg \gg \left(2e^{i\frac{\pi }{2}}\right)\cdot \left(5e^{i\frac{\pi }{3}}\right)\\ \\ \gg \gg (2\cdot 5)\left(e^{i\frac{\pi }{2}}\right)\left(e^{i\frac{\pi }{3}}\right)\gg \gg 10e^{i\frac{\pi }{2}+i\frac{\pi }{3}}\gg \gg 10e^{i\frac{5\pi }{6}}\end{array}$$

Quotient $z/w$:

$$\begin{array}{c}z/w\gg \gg \left(2e^{i\frac{\pi }{2}}\right)/\left(5e^{i\frac{\pi }{3}}\right)\\ \\ \gg \gg \frac{2e^{i\frac{\pi }{2}}}{5e^{i\frac{\pi }{3}}}\gg \gg \frac{2}{5}{e}^{i\frac{\pi }{2}}{e}^{-i\frac{\pi }{3}}\gg \gg \frac{2}{5}{e}^{i\frac{\pi }{6}}\end{array}$$

Power $z^8$:

$$\begin{array}{c}{z}^8\gg \gg {\left(2e^{i\frac{\pi }{2}}\right)}^8\\ \\ \gg \gg 2^8{\left(e^{i\frac{\pi }{2}}\right)}^8\gg \gg 512e^{i\cdot 4\pi }\end{array}$$

Notice:

$$e^{i\cdot 4\pi }\gg \gg {\left(e^{2\pi i}\right)}^2\gg \gg 1^2\gg \gg 1$$

Simplify:

$$512e^{i\cdot 4\pi }\gg \gg 512$$ Link to original

Example - Complex power from Cartesian

Complex power from Cartesian

Compute ${(3+3i)}^4$.

Solution

First convert to exponential form:

$$\begin{array}{c}3+3i\gg \gg 3\sqrt{2}(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)\\ \\ \gg \gg 3\sqrt{2}{e}^{i\frac{\pi }{4}}\end{array}$$

Compute the power:

$$\begin{array}{c}{(3+3i)}^4\gg \gg {\left(3\sqrt{2}{e}^{i\frac{\pi }{4}}\right)}^4\\ \\ \gg \gg 324e^{i\pi }\gg \gg -324\end{array}$$Link to original

Complex roots

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06 Theory - Roots formula

Theory 1

The exponential notation leads to a formula for a complex $n^{\text{th}}$ root of any complex number:

$$\sqrt[n]{r{e}^{i\theta }}=\sqrt[n]{r}{e}^{i\frac{\theta }{n}}$$

$n$ distinct roots

Every complex number actually has $n$ distinct $n^{\text{th}}$ roots!

That’s two square roots, three cube roots, four $4^{\text{th}}$ roots, etc.

All complex roots

The complex roots of $z=r{e}^{i\theta }$ are given by this formula:

$$w_k=\sqrt[n]{r}\cdot e^{i(\frac{\theta }{n}+k\frac{2\pi }{n})}\text{for each }k=0,1,2,\dots ,n-1$$

In Cartesian notation:

$$w_k=\sqrt[n]{r}\cos (\frac{\theta }{n}+k\frac{2\pi }{n})+\sqrt[n]{r}\sin (\frac{\theta }{n}+k\frac{2\pi }{n})i$$

In words:

• Start with the basic root: $\sqrt[n]{r}\cdot e^{i\frac{\theta }{n}}$
• Rotate by increments of $\frac{2\pi }{n}$ to get all other roots
  • After $n$ distinct roots, this process repeats itself

Extra - Complex roots proof

We must verify that $w_k^n=r{e}^{i\theta }$:

$$\begin{array}{c}{(\sqrt[n]{r}\cdot e^{i(\frac{\theta }{n}+k\frac{2\pi }{n})})}^n\gg \gg r^{\frac{n}{n}}\cdot e^{i(\frac{\theta }{n}+k\frac{2\pi }{n})n}\\ \\ \gg \gg r\cdot e^{i(\theta +2\pi k)}\gg \gg r{e}^{i\theta }{e}^{i2\pi k}\gg \gg r{e}^{i\theta }\end{array}$$

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07 Illustration

Example - Finding all $4^{\text{th}}$ roots of $16$

Finding all 4th roots of 16

Compute all the $4^{\text{th}}$ roots of $16$.

Solution

Write $16=16e^{0i}$.

Evaluate roots formula:

$${\left(16e^{0i}\right)}^{\frac{1}{4}}\gg \gg w_k={16}^{\frac{1}{4}}{e}^{i(\frac{0}{4}+k\frac{2\pi }{4})}$$

Simplify:

$$\gg \gg 2e^{i\cdot k\frac{\pi }{2}}\gg \gg 2,2i,-2,-2i$$ Link to original

Example - Finding $2^{\text{nd}}$ roots of $2i$

Finding 2nd roots of 2i

Find both $2^{\text{nd}}$ roots of $2i$.

Solution

Write $2i=2e^{i\frac{\pi }{2}}$.

Evaluate roots formula:

$$\begin{array}{c}{\left(2e^{i\frac{\pi }{2}}\right)}^{\frac{1}{2}}\gg \gg w_k=\sqrt{2}{e}^{i(\frac{\pi /2}{2}+k\frac{2\pi }{2})}\\ \\ \gg \gg \sqrt{2}{e}^{i(\frac{\pi }{4}+k\pi )}\end{array}$$

Compute the options: $k=0,1$:

$$\gg \gg \sqrt{2}{e}^{i\frac{\pi }{4}},\sqrt{2}{e}^{i\frac{5\pi }{4}}$$

Convert to rectangular:

$$\begin{array}{c}\gg \gg \sqrt{2}(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i),\sqrt{2}(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i)\\ \\ \gg \gg 1+i,1-i\end{array}$$ Link to original

Example - Some roots of unity

Some roots of unity

Find the $1^{\text{st}}$ and $2^{\text{nd}}$ and $3^{\text{rd}}$ and $4^{\text{th}}$ and $5^{\text{th}}$ and $6^{\text{th}}$ roots of the number $1$.

Solution

(1) $1^{\text{st}}$

Write $1=e^{0i}$. Evaluate roots formula. There is no possible $k$:

$${\left(e^{0i}\right)}^{\frac{1}{1}}\gg \gg e^{0i}\gg \gg 1$$

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(2) $2^{\text{nd}}$

Write $1=e^{0i}$. Evaluate roots formula in terms of $k$:

$${\left(e^{0i}\right)}^{\frac{1}{2}}\gg \gg w_k=e^{i(\frac{0}{2}+k\frac{2\pi }{2})}k=0,1$$

Compute the two options, $k=0,1$:

$$\gg \gg 1,e^{\pi i}\gg \gg 1,-1$$

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(3) $3^{\text{rd}}$

Evaluate roots formula in terms of $k$:

$${\left(e^{0i}\right)}^{\frac{1}{3}}\gg \gg w_k=e^{i(\frac{0}{3}+k\frac{2\pi }{3})}$$

Compute the options: $k=0,1,2$:

$$\gg \gg 1,e^{i\frac{2\pi }{3}},e^{i\frac{4\pi }{3}}\gg \gg 1,-\frac{1}{2}+\frac{\sqrt{3}}{2}i,-\frac{1}{2}-\frac{\sqrt{3}}{2}i$$

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(4) $4^{\text{th}}$

Evaluate roots formula:

$${\left(e^{0i}\right)}^{\frac{1}{4}}\gg \gg w_k=e^{i(\frac{0}{4}+k\frac{2\pi }{4})}$$

Compute the options: $k=0,1,2,3$:

$$1,e^{i\frac{\pi }{2}},e^{i\pi },e^{i\frac{3\pi }{2}}\gg \gg 1,i,-1,-i$$

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(5) $5^{\text{th}}$

Evaluate roots formula:

$${\left(e^{0i}\right)}^{\frac{1}{5}}\gg \gg w_k=e^{i(\frac{0}{5}+k\frac{2\pi }{5})}$$

Compute the options: $k=0,1,2,3,4$:

$$1,e^{i\frac{2\pi }{5}},e^{i\frac{4\pi }{5}},e^{i\frac{6\pi }{5}},e^{i\frac{8\pi }{5}}$$

Don’t simplify, it’s not feasible.

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(6) $6^{\text{th}}$

Evaluate roots formula:

$${\left(e^{0i}\right)}^{\frac{1}{6}}\gg \gg w_k=e^{i(\frac{0}{6}+k\frac{2\pi }{6})}$$

Compute the options: $k=0,1,2,3,4,5$:

$$1,e^{i\frac{2\pi }{6}},e^{i\frac{4\pi }{6}},e^{i\frac{6\pi }{6}},e^{i\frac{8\pi }{6}},e^{i\frac{10\pi }{6}}$$

Simplify:

$$\gg \gg 1,\frac{1}{2}+\frac{\sqrt{3}}{2}i,-\frac{1}{2}+\frac{\sqrt{3}}{2}i,-1,-\frac{1}{2}-\frac{\sqrt{3}}{2}i,\frac{1}{2}-\frac{\sqrt{3}}{2}i$$Link to original