MandysNotes

The real part of \[ e^{i\alpha} \ \]

is:

\[ \text{Re}(e^{i \alpha}) = \frac{ e^{i\alpha} + e^{-i\alpha}}{2} \]

\[ =\sum_{k=0}^{\infty} (-1)^{k}\frac{\alpha^{2k}}{(2k)!} . \]

 

The imaginary part of \[ e^{i\alpha} \ \]

is:

\[ \text{Im}(e^{i\alpha}) = \frac{ e^{i\alpha} - e^{-i\alpha}}{2i} \]

\[ = \sum_{k=1}^{\infty}(-1)^{k -1}\frac{\alpha^{2k -1}}{(2k -1)!} . \]

 

We define:

\[ \text{Re}(e^{i\alpha}) = \frac{ e^{i\alpha} + e^{-i\alpha}}{2} := \cos{\alpha} ;\]

\[ \text{Im}(e^{i\alpha}) = \frac{ e^{i\alpha} - e^{-i\alpha}}{2i} := \sin{\alpha}.\]

 

Therefore:

\[ e^{i\alpha} = \text{Re} (e^{i\alpha}) + i\text{Im}(e^{i\alpha}) =\cos{\alpha} + i \sin{\alpha}. \]

 

This is Euler's Formula.

01 May 2011 In Complex Numbers

For any complex number \[ z = x + iy ,\]

\[ z \bar{z} = (x + iy)(x -iy) \]

\[ = x^{2}  + ixy - ixy + y^{2} =   x^{2} + y^{2} .\]

That is, any complex number times its complex conjugate equals the square of its real part plus the square of its imaginary part.

Because both the real and imaginary parts of any complex number are real:

\[ z \bar{z} = x^{2} + y^{2} \geq 0,   \forall z \in \mathbb{C}.\]

We define the positive square root of \[ z \bar{z}: \]

\[ +\sqrt{z\bar{z}} : = |z| ,\]

to be the Modulus of the complex number \[ z .\]

01 May 2011 In Complex Numbers

Recall that a polynomial of degree two is called a Quadratic Polynomial.

A general quadratic polynomial (over the real numbers) is of the form:

\[ az^{2} + bz + c = 0. \]

The formula for solving for the roots of this equation is:

\[ z = \frac{-b \pm \sqrt{b^{2} -4ac}}{2a} .\]

01 May 2011 In Complex Numbers

 

Take a pencil and place it along the edge of a ruler, so that the ruler and the pencil are parallel, and so that the end of the eraser is at zero, and the point of the pencil points towards increasing numbers on the ruler.

RulerPencilH

 

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NOTRad