Complex/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.
Comments Left
how do i simply multiply imaginary numbers?
like;
(-4i)(-5i)(3i)
What do you do if you are given the relationship between two complex numbers and the equation of the locus of one, and you are asked to find the locus of the other? For example
u+iv=(x+iy)^2
y=x-1
Find the locus of u+iv
@Jack
remember that the i makes the y terms squared negative. Now we substitute y=x-1 into this equation, since that is what the other complex number satisfies, so we have an equation in terms of u,v and x.This gives:
Your basically looking for a equation containing only u and v so we need to eliminate the x and y. Firstly take the equation you have with u and v in it and expand it out to get
$latex u + iv = (2x^2 -2)i + 2x -1
We now can compare the real and imaginary terms of each side of the equation to get
and
Substituting the equation for x in terms of u into the equation for v get
which you can simplify to get
You may need to check my algebra but the basic method is:
>get a simple equation in terms of u,v and x or y or both
>use the locus of the other complex number to find an expression for either x or y
>substitute this into you equation containing u and v so that this only has one other variable, say x
>compare real and imaginary parts to gain two equations of real numbers in terms of u,v and x
>use these two to eliminate x (make x the subject of one and substitute into the other)
>simplify the resulting expression
you can treat them like you would in normal algebra but whenever you get a i squared you can chnage it to -1
ie in your example the answer would be
(-4i)(-5i)(3i) = -60i
beacuse 4x5x3 = 20 and i x i x i = -i
hope that helps dave