# Complex Numbers: The Imaginary Number

#### The Imaginary Number

Numbers that have real and imaginary parts are complex numbers. They can be expressed in the form $a+b i$ where i is $\sqrt{-1}$ where $a,b\in\mathbb{R}$. Real numbers are actually a special case of complex numbers. Take the above expression. If b is zero then there is no imaginary part and the number is real. As with all special cases the original case has properties that are very much different for example, complex conjugates: Let   $\widetilde{z}=a+bi$   (where   $\widetilde{ }$   denotes that z is a complex number) then the complex conjugate of   $\widetilde{z}=a+bi$   is

$\widetilde{z}^*=\overline{\widetilde{z}}=a-bi$.  This is also true for numbers such as  $\widetilde{z}=A \sin (i \theta )$, here the complex conjugate is   $\overline{\widetilde{z}}=\overline{A} \sin (-i\theta)$.  The complex conjugate is generally signified by either the asterisk $z^*$ or the bar above the number $\overline{z}$. Since real numbers have no complex part the complex conjugate of a real number is itself i.e.   $\overline{z}=z$.

Before moving forward let’s consider a complex number of the form $\frac{a_1}{a_1+ib_1}$. This number can be rewritten by multiplying by one as follows

$\frac{a_1}{a_1+ib_1}=\frac{a_1}{a_1+ib_1}\times\frac{a_1-ib_1}{a_1-ib_1}=\frac{a_1\times (a_1-ib_1)}{a_{1}^2+b_{1}^{2}}=\frac{a_{1}^{2}-ia_{1}b_{1}}{a_{1}^{2}+b_{1}^{2}}$

Complex numbers can be viewed as vectors in the complex plane as shown below:

Math and physics use $i$ as the symbol for $\sqrt{-1}$ but many other fields including engineering use $j$ instead. One reason for this may be that engineers use the letter i to represent current and so the next letter j became the symbol for the imaginary number.

Among some of the most useful equations to use with imaginary numbers are the following formulas:

$\sin (\theta ) = \frac{e^{i\theta} - e^{-i\theta } }{2i}$

$\sin (\theta ) = \frac{e^{i\theta} + e^{-i\theta } }{2}$

Notice that if we plug in a+bi then we see that the sine component of the imaginary number is b and the cosine part of is a.

While we didn’t talk about it, De Movier’s Theorem can be very useful, and for information on De Movier’s Theorem (and for more on complex numbers) Dr. Schneider at Portland State University wrote a very article here.

#### Derivatives of Complex Functions

While complex analysis can easily become very difficult, here we shall look through it at a beginner’s level.

Hopefully, you’ve had some experienced some calculus. If not, then this section may be a little difficult to understand.

Now a function is differentiable at a point, c, if $\lim_{h\to 0 } \frac{f(c+h) - f(c)}{h}$ exists. Also, the function must be holomorphic (or an analytic) in a specified region (holomorphic implies that the function is infinitely differentiable).

This brings us to one of the interesting facts discovered by Cauchy and Riemann: If a function $f(z) = u + i v$ is complex differentiable, then its real and imaginary parts satisfy the Cauchy-Riemann equations:

$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$

and

$\frac{\partial v}{\partial x} = - \frac{\partial u}{\partial y}$

Where $x=Re[z]$, $y=Im[z]$, $u=Re[f(z)]$, and $v=Im[f(z)]$

Well, with that let’s practice this!

Find the real and imaginary part of the following expressions:

1. $(3i+1)^2 [2+(4-i)^2]$

$(-9+3i+3i+1)(2+16-4i-4i-1)=(-8+6i)(17-i)=-136+64i+102i+48=-88+166i$

Real: -88

Imaginary: 166

2.      $2i\frac{(4+5i)(2-3i)}{(2+3i)^2}$

$2i\frac{8-12i+10i+15}{4+6i+6i-9}=2i\frac{23-2i}{12i-5}=\frac{4+46i}{12i-5}\times\frac{-5-12i}{-5-12i}=\frac{-20-48i-23-i+552}{25+144}=\frac{532-278i}{169}$

Real: $\frac{532}{169}\approx 3.15$

Imaginary: $-\frac{278}{169}\approx -1.64$

Plot the values of $4e^{i\theta}$ for $\theta = 0,\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi }{2},\pi$

Prove the magnitude of a complex number, z, is equal to $\sqrt{z z^*}$

Let $z\equiv a + i b$, $a,b\in\mathbb{R}$. Then $\sqrt{z z^*}=\sqrt{(a-ib)(a+ib)}=\sqrt{a^2+iab-iab+b^2}=\sqrt{a^2+b^2}=|z|$.

Prove that $Re[z] = \frac{z+z^*}{2}$ and that $Im[z]=\frac{z-z^*}{2}$

$\frac{z+z^*}{2}=\frac{a+ib+a-ib}{2}=a=Re[z]$

$\frac{z-z^*}{2}=\frac{a+ib-(a-ib)}{2i}=b=Im[z]$

Prove De Moivre’s Theorem. In other words, prove that if $r\equiv |z|$ is the magnitude of z and $\theta$ is the angle between the real axis and the line that goes from (0,0) to z in an Argand Diagram, then $z^n=r^ne^{in\theta}=r^n(cos(n\theta )+isin(n\theta ))$

Let $z\equiv re^{i\theta }$. Then $z^n=(r e^{i\theta })^n=r^ne^{in\theta}=r^n(cos(n\theta )+isin(n\theta ))$

Use Euler’s formula to prove that $sin(a+b)=cos(a)sin(b)+sin(a)cos(b)$

If we were to first look at the exponential form of these two equaions we would see that $e^{i(a+b)}=e^{ia}e^{ib}$. Now, $cos(a+b)+isin(a+b)=[cos(a)+isin(a)][cos(b)+isin(b)]=cos(a)cos(b)-sin(a)sin(b)+i(cos(a)sin(b)+cos(b)sin(a))$. For this to be true, the real part on the left has to equal the real part of the equation on the right. So, we see the following identities:

1. $sin(a+b)=cos(a)sin(b)+sin(a)cos(b)$
2. $cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$

Use Euler’s formula to write $sin(\theta )$ and $cos(\theta )$ in terms of exponentials

$e^{i\theta }+e^{-i\theta } = cos(\theta ) + isin(\theta ) + cos(-\theta ) +isin(-\theta )=2cos(\theta )\implies \frac{e^{i\theta}+e^{-i\theta}}{2}$

$e^{i\theta }-e^{-i\theta } = cos(\theta ) + isin(\theta ) - cos(-\theta ) -isin(-\theta )=2isin(\theta )\implies \frac{e^{i\theta}-e^{-i\theta}}{2i}$

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