Complex and Imaginary Numbers

Most of the time in maths you are working with “real” numbers ie) all the rational numbers so we have 1,3,3.5567,4/7,root 2 etc, however sometimes we will need to extend our field of numbers to imaginary numbers. This is when we want to find the root of a negative number eg root -9, as (-3)² and 3² both equal 9 we are unable to find the square root of -9. In order to solve this number we can use the “imaginary” number i such that

i = √-1

We then say that √-9 = 3i since √ab = √a √b and therefore √-9=√9 √-1 = 3i . This allows us to solve all equations with negative roots. It is also useful to know what iⁿ gives as shown below

i² = -1

i³=-i

i⁴=1

i⁵=i

Complex numbers are made up of both real and imaginary parts,

eg z =a+ib

is a complex number. When comparing complex numbers we need only compare the real and imaginary parts, by equation them to each other,

ie) if the complex number z=a+ib

and 2z = 6+10i

a=3 and b=5

To solve complex equations with fractions we often have to rationalise the fractions, like we would with surd’s, so that there are no imaginary parts on the bottom and we can compare the coefficient’s, this is done by multiplying by the conjugate.

note// the conjugate of z=a+ib is z’ = a-ib

eg) (2+3i)/(4+5i) = (2+3i)(4-5i)/(4+5i)(4-5i) = (23+2i)/41

Complex numbers can be represented on an argand diagram when the y axis is the imaginary part and the x axis is the real part, so that the number a+ib is plotted at (a,b), and a line is draw between this and the origin to represent the number. The length of the line is called the moduls and the angle between the real axis and the line is the argument.

Trigonometry Identities

There a number of “identities” in trigonometry that can be found from the basic ideas of sin, cos and tan as explained in my earlier post. These identities can help in solving equations involving trig functions, especially when there are 2 or more different functions as the often allow you to write the equation in terms of one function, eg sin, that you can then solve.

One of the identities is:

sin² + cos² = 1.

To prove this consider a right angled triangle with side a,b and c as shown below

From this we can use Pythagoras theorem to say:

a²+b²=c²

now we know

sin t = b/c so b = csin t

cos t = a/c so a = ccos t

substituting these values in the above equation we get

(csint)² +(ccost)² = c²

canceling the c² we get

sint² + cost² = 1

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There are trig functions that are equal to 1 over sin, cos and tan called cosec = 1/sin, sec = 1/cos and cot = 1/tan. These can be remembered using the third letter rule as the third letter of each of these corresponds to the the function it is one over.

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Using these a cos² + sin² = 1 we can calculate other identities

tan²t + 1 = sec^2t

We can obtain this by dividing through by cos² as we know sin/cos = tan, cos/cos = 1 and 1/cos = sec.

Other similar identities can be obtained for cosec and cot.

Hyperbolic Functions

Hyperbolic functions are similar to sin,cos tan etc in trigonometry and share many similar rules. Usually hyperbolic functions are written like the trigonometric ones but with a h on the end, eg sinh and cosh.

The hyperbolic functions can be all written in terms of e, sinh and cosh are as follows

sinh(x) = (e^x - e^-x)/2
cosh(x) = (e^x + e^-x)/2

And tanh can be defined as sinh/cosh so

tanh = (e^x - e^-x) / (e^x + e^-x)

though this is often written as
tanh = (e^2x - 1) / (e^2x + 1)
by timesing the top and bottom by e^x

the other other hyperbolic functions sinh as sech, coth etc can be found in the same way as they would be in trigonometry, by using 1 over the other functions, ie sech = 1/cosh

Most of the identities in trigonometry have a similar identity with hyperbolic functions, however in most of these whenever there is a sin² it changes to a -sinh²
so
cosh² - sinh²=1
which you can work out by placing the equations with e’s in the place of sinh and cosh