Question: What is the easiest way to understand complex numbers and their forms?

[Xmo91]

Homework Statement

Im am having trouble grasping the concept of complex numbers, could anyone please take the time to explain it. I have asked my lecturer but his help just seems to confuse me more. I have also searched the internet looking for explanations but that is giving me mixed messages.

Homework Equations

The equation that I am currently working on is probably really easy but could anyone explain it?

z^2 + 4iz-3 =0

The Attempt at a Solution

 

[yyat]

Hi Xmo91!

The equation is a quadratic equation in z, it has two solutions, which are both complex numbers in this case. You can solve it with the usual formula for quadratic equations.

There are several ways to define complex numbers, but all equivalent, which may be the reason you got "mixed messages" (could you be more specific?).

Complex numbers are just pairs (a,b) of real numbers, which are usually written as a+bi (a is the real part, b is the imaginary part). To compute with complex numbers you can use the usual properties from the real numbers, the only new thing you need to know is that i^2=-1. Complex numbers take some getting used to, but once you have mastered the basics, they are really not much more difficult then the real numbers to work with.
There is of course a lot more to say about complex numbers, but I could only repeat what you find in other internet resources, for example http://en.wikipedia.org/wiki/Complex_numbers" .

I suggest you take a look at some of the available texts and ask more specific questions if there is something you don't understand.

 

[tiny-tim]

Welcome to PF!

Hi Xmo91! Welcome to PF!

Complex numbers generally work the same way as real numbers …

the only difficulty usually is when you have a fraction, like (a +ib)/(c + id), which you deal with by the trick of multiplying both top and bottom by (c - id), to make the bottom real …

try solving your problem …

z^2 + 4iz-3 =0

… just like any ordinary quadratic equation:

what do you get? ​

 

[[Quadratic]]

To expand a bit on something tiny-tim mentioned, anytime you have a complex number a+bi and you multiply by that number's conjugate a-bi, you always end up with a real number n+0i which equals n.

ex. (2+3i)(2-3i)=4-6i+6i-9(i²)
=4-9(-1)
=4-(-9)
=4+9
=13

 

[Xmo91]

I got a similar eqn in class so i worked through it and got this:

Z^2 + 2iz -4=0

x=(-b+/- sqrt(b^2 - 4ac))/2a

where a=1 b=2i and c=-4

x=[-2i+/- sqrt(2i^2 - 4x1x-4)]/2x1
x=[-2i+/- sqrt(2(-1) - (-16))]/2
x=[-2i+/- sqrt(-2 - (-16))]/2
x=[-2i+/- sqrt(14)]/2
x=[-2i+3.741]/2 or x=[-2i-3.741]/2
at this stage i was alittle confused as to what was my next step, so i just simplified my answer
x=-i+1.8705 or x=-i-1.8705

 

[yyat]

x=[-2i+/- sqrt(2i^2 - 4x1x-4)]/2x1

should be (2i)^2, not 2i^2.

 

[tiny-tim]

Hi Xmo91!

(have a square-root: √ and try using the X² tag just above the Reply box )

at this stage i was alittle confused as to what was my next step, so i just simplified my answer
x=-i+1.8705 or x=-i-1.8705

Yes, and then the usual way of writing the result would be x = -i ± 1.8705 …

except of course, as yyat points out, you need to re-do that square-root

 

[Xmo91]

Thanks for the help guys, just one more thing before this thread dies, can anyone please explain what the following question is asking me to do?

Find the values of z that satisfy z^3 = √i and then I'm asked to express it in Cartesian form, which i know has to do with that circle diagram (the name slips my mind).

 

[HallsofIvy]

Thanks for the help guys, just one more thing before this thread dies, can anyone please explain what the following question is asking me to do?

Find the values of z that satisfy z^3 = √i and then I'm asked to express it in Cartesian form, which i know has to do with that circle diagram (the name slips my mind).

It's a little hard to answer because you haven't told us what you DO know. since you mention "Cartesian form", a+ bi, do you know the "polar form", $r(cos(\theta)+ i sin(\theta))$? Do you know that can be written in "exponential form", $r e^{i\theta}$?

It you know polar form then you probably also know DeMoivre's theorem that says that
$$[r(cos(\theta)+ i sin(\theta))]^n= r^n(cos(n\theta)+ i sin(n\theta))$$

That applies for any real n, including fractions so
[tex][r (cos(\theta)+ i sin(\theta))]^{1/n}= r^{1/n}(cos(\theta/n)+ i sin(\theta/n))[/itex]<br /> and by adding $2\pi i$ repeatedly to $\theta$ you can get all n roots.<br /> <br /> i itself can be written $1(cos(\pi/2)+ i sin(\pi/2)$ since $cos(\pi/2)= 0$ and $sin(\pi/2)= 1$. So $\sqrt{i}= cos(\pi/4)+ i sin(\pi/4)$. You can use DeMoivre's theorem to find the three cube roots of that.<br /> <br /> <br /> (There is an ambiguity here. If the problem had be to solve z⁶= i, there would be 6 roots because i is the square of both $cos(\pi/4)+ i sin(\pi/4)$ and $-cos(\pi/4)- i sin(\pi/4)$. Here I have use the real number convention that $\sqrt{}$ refers only to the "positive" root but, since the complex numbers do not form an ordered field, that distinction is not really valid.)[/tex]

 

[tiny-tim]

Find the values of z that satisfy z^3 = √i and then I'm asked to express it in Cartesian form, which i know has to do with that circle diagram (the name slips my mind).

Quick answer …

Polar form: re^iθ (that's your "circle diagram" )

Cartesian form: x + iy​

Cartesian form is easier for most questions, but polar form is easier for questions like z³ = √i

The names are the same as for coordinates:

Polar coordinates: (r,θ)
Cartesian coordinates: (x,y)​