3-ComplexNumSummary

= Complex Numbers : Summary =


 * Complex Numbers **

A complex number is usually written in the form: z = x + iy, where math . \quad \;. \;\; i = \sqrt{-1} math
 * x is the Real part of z
 * Re(z) = x
 * x is a real value
 * y is the Imaginary part of z
 * Im(z) = y
 * y is a real value (the coefficient of i)

z = 2 – 4i
 * Example **


 * Re(z) = 2
 * Im(z) = –4


 * Argand Diagram **

Is a set of axes with a Real Axis and an Imaginary Axis.

A complex number is a single point on the Argand Diagram.


 * Addition & Subtraction **

To add and subtract complex numbers
 * add or subtract the real parts together
 * add or subtract the imaginary parts together


 * Example **

... ... (2 + 3i) + (5 + i) = 7 + 4i

... ... (6 – 2i) – (3 + 2i) = 3 – 4i


 * Multiplication **

To multiply complex numbers, write the number in brackets and expand the brackets as per basic algebra.

Don't forget: ...... i 2 = –1


 * Example **

... ... 5(2 + 3i) = 10 + 15i

... ... (2 + 3i)(4 – 2i) = 8 – 4i + 12i – 6i 2 ... ... ... ... ... ... ... ... ... = 8 +8i + 6 ... ... ... ... ... ... ... .... ... = 14 + 8i


 * Division **

Recall that the conjugate of z is the result of reversing the sign for the imaginary part of z.

math \\ . \qquad z = 3 + 2i \\. \\ . \qquad \overline{z} = 3 - 2i math

Multiply by the conjugate of the denominator over itself


 * Example **

math \\ . \qquad \dfrac{2 + 3i}{4 - 2i} = \dfrac{2 + 3i}{4 - 2i} \times \dfrac{4 + 2i}{4 + 2i} \\. \\ . \qquad \qquad \qquad = \dfrac{ (2+3i)(4 + 2i) }{ (4 - 2i)(4 + 2i) } \\. \\ . \qquad \qquad \qquad = \dfrac{ 8 + 4i + 12i +6i^2 }{ 16 + 8i - 8i - 4i^2 } math . math \\ . \qquad \qquad \qquad = \dfrac{ 8 + 16i - 6}{16 + 4} \\. \\ . \qquad \qquad \qquad = \dfrac{2 + 16i}{20} \\. \\ . \qquad \qquad \qquad = \dfrac{2}{20} + \dfrac{16i}{20} math . math . \qquad \qquad \qquad = \dfrac{1}{10} + \dfrac{4i}{5} math

If we draw a line on the argand diagram from the Origin to a Point z = x + yi
 * Polar Form **

The Modulus of z is the length of that line

math . \qquad r = \text{Mod}(z) = |z| = \sqrt{x^2 + y^2} math

The Argument of z is the angle to the positive real axis. The Argument **__must__** be expressed as an angle between –p and p

math . \qquad \theta = \text{Arg}(z) = \text{Tan}^{-1} \left( \dfrac{y}{x} \right) \quad \theta \in (-\pi, \;\pi] math

Using basic trig, we get:
 * x = rcos(q )
 * y = rsin(q )

So
 * z = rcos(q ) + irsin(q )
 * z = rcis(q )


 * Example **

math \text{Write } \; z = -5\sqrt{3} - 5i \; \text{ in Polar Form} math


 * Solution: **

math \\ . \qquad r = \sqrt{ \left( -5 \sqrt{3} \right)^2 + \big( -5 \big)^2 } \\. \\ . \qquad \;\;\; = \sqrt{ 75 + 25 } \\. \\ . \qquad \;\;\; = \sqrt{100} \\. \\ . \qquad \;\;\; = 10 math . math \\ . \qquad \theta = \text{Tan}^{-1} \left( \dfrac{ -5 }{ -5\sqrt{3} } \right) \\. \\ . \qquad \;\;\; = \text{Tan}^{-1} \left( \dfrac{1}{\sqrt{3}} \right) \\. \\ . \qquad \;\;\; = \dfrac{\pi}{6} math {but z is in Quadrant 3 so so change q but keep in domain (–p, p ] }

math . \qquad \theta = - \dfrac{5\pi}{6} math

Hence math . \qquad z = 10\, \text{cis} \left( - \dfrac{5\pi}{6} \right) math

To change from Polar Form back to Rectangular Form, remember that:
 * x = rcos(q )
 * y = rsin(q )


 * Multiplying Numbers in Polar Form **

math \text {For } z_1 =r_1 \, \text {cis} \, \theta_1 \; \text { and } \; z_2 = r_2 \, \text{cis} \, \theta_2 math

math z_1 \times z_2 = \left( r_1r_2 \right) \, \text {cis} \left( \theta_1 + \theta_2 \right) math


 * Example **

math 2 \, \text{cis} \left( \dfrac{\pi}{4} \right) \times 3 \, \text{cis} \left( \dfrac{\pi}{3} \right) = 6 \, \text{cis} \left( \dfrac{7\pi}{12} \right) math

Don't forget the Principal Domain for q : (-p, p] . If your answer gives an angle outside of that domain, you must change it.


 * Dividing Numbers in Polar Form **

Same idea as multiplying (divide the r values, subtract the angles)


 * DeMoivre's Theorem **

math \text {For } z = r \text{cis} \theta math

De Moivre's Theorem gives us the nth power of z math z^n = r^n \, \text{cis} (n\theta) \quad n \in Q \text { (Rational numbers)} math


 * Example **

math \text{Find } \; z^3 \quad \text{given } \; z = 2 \, \text{cis} \left( \dfrac{2\pi}{5} \right) math

math \\ . \qquad z^3 = 2^3 \, \text{cis} \left( 3 \times \dfrac{2\pi}{5} \right) \\. \\ . \qquad \;\;\; = 8 \, \text{cis} \left( \dfrac{6\pi}{5} \right) \\. \\ . \qquad \;\;\; = 8 \, \text{cis} \left( -\dfrac{4\pi}{5} \right) math


 * Solving Equations in the Complex Number Field **

Any equation of degree n, when solved in the complex number field will have n linear solutions.
 * some of those linear solutions may be repeated (ie a squared term)

When solving any equation with all __**real**__ coefficients, any complex solutions will appear in conjugate pairs.

A quadratic equation can be solved over C in two ways.
 * Using the Quadratic Formula
 * replace any negative value inside the square root with i 2.
 * By completing the square
 * if you end up with the sum of two squared terms, then replace the second term with –i 2.
 * then factorise using the difference of two squares method.


 * Example 1 **

math \text {Solve } \; 2z^2-2z+3=0 math

{Use Quadratic Formula with a = 2, b = –2, c = 3} math \\ . \qquad z=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \\. \\ . \qquad z=\dfrac{2 \pm \sqrt{(-2)^2 - 4 \times 2 \times 3}}{2 \times 2} \\. \\ . \qquad z=\dfrac {2 \pm \sqrt{-20}}{4} math

{Now simplify the surd, introduce i and separate fraction into real and imaginary parts} math \\ . \qquad z=\dfrac {2 \pm \sqrt{20i^2}}{4} \\. \\ . \qquad z=\dfrac {2}{4} \pm \dfrac {2 \sqrt{5} \, i}{4} \\. \\ . \qquad z=\dfrac {1}{2} \pm \dfrac {\sqrt{5} \, i}{2} math


 * Example 2 **

math \text{Solve } \; z^2+6z+13 = 0 \quad z\in C math

Factorise by completing the square

math \\ . \qquad z^2+6z+13 = 0\\. \\ . \qquad \underline{z^2+6z+9} \; -9+13 = 0\\. \\ . \qquad (z+3)^2+4 = 0 math

In the real field, this can't be factorised. But introducing i 2 = –1 makes this a difference of two squares

math \\ . \qquad (z+3)^2-4i^2 = 0 \\. \\ . \qquad (z+3+2i)(z+3-2i) = 0 math

math \\ . \qquad z = -3 - 2i \;\text{ or } \; z = -3 + 2i \\ \text{or} \\ . \qquad z = -3 \pm 2i math

Using DeMoire to Find square (or higher) roots of complex numbers

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