COMPLEX NUMBERS

Euler’s relation

e^iθ = cosθ + isinθ

Euler’s relation in exponential form

z = re^iθ

where r is the magnitude of z and θ is the argument

x = rcosθ

y = rsinθ

e^-iθ

cosθ - isinθ

|z1||z2|

|z1z2|

arg(z1z2)

arg(z1) + arg(z2)

|z1/z2|

|z1| / |z2|

arg(z1/z2)

arg(z1) - arg(z2)

de Moivre’s theorem

z^n = r^n(cosnθ + isinnθ)

or (r(cosθ + sinθ))^n

cosθ - isinθ (not exponenetial)

cos(-θ) + isin(-θ)

proving that cos or sinnθ = a crazy equation

• use de Moivre’s to say that (cosθ + isinθ)^n = cosnθ + isinnθ

• then, solve the term with the index in a binomial expansion

• depending on whether you’re solving for cos or sin, eliminate all the imaginary or real terms respectively and equate to either sin or cos.

• it will expand to the correct statement

z + 1/z

2cosθ

z - 1/z

2isinθ

z^n + 1/z^n

2cosnθ

z^n - 1/z^n

2isinnθ

proving and index term (cos^nθ) equals an equation where the n in in front of θ e.g. cos5θ

• for the example on the front of the card, it would be that 2cosθ = z + 1/z and thus:

• 32cos^5θ = (z + 1/z)^5

• then expand the brackets with the z and at the end sub the trigonometric back in and solve as normal

sum of a series with n-1 on top

Σwz^r = w(z^n - 1) / z-1

where:

w = first term

z = a complex number of phrase

n = the index (or the number on top of sigma +1)

sum of a series to infinity

Σwz^r = w / 1-z^r where |z| < 1

de Moivre’s theorem for geometric problems / nth roots of a complex number

z = cosθ + isinθ = cos(θ + 2kπ) + isin(θ + 2kπ)

how to solve z^n = a general term

• put into terms of cos and sin

• then say z = r(cosθ + isinθ)

• thus, z^n = r^n(cosnθ + isinnθ)

• equate this to the term given in the question with the +2kπ and substitute in n different values of k to find n roots but keep the argument between -π and π

what do the roots of z^n look like on an Argand diagram

the points will turn out to be a regular polygon, where each point is an equal angle along from the last point around the origin

what is the sum of roots of z^n = 1?

0

solving geometric problems

• the only thing that’s going to change between different roots is the argument.

• therefore, as it will form a polygon around the origin, do the 2kπ again, except this time, it will by 2kπ/n