# Homework Help: Complex Numbers concept help

1. Mar 16, 2009

### Xmo91

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
The equation that im currently working on is probably really easy but could anyone explain it?

z^2 + 4iz-3 =0

3. The attempt at a solution

Last edited: Mar 16, 2009
2. Mar 16, 2009

### 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" [Broken].

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.

Last edited by a moderator: May 4, 2017
3. Mar 16, 2009

### 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 …

… just like any ordinary quadratic equation:

what do you get?

4. Mar 16, 2009

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

Last edited: Mar 17, 2009
5. Mar 17, 2009

### 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

6. Mar 17, 2009

### yyat

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

7. Mar 17, 2009

### tiny-tim

Hi Xmo91!

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

8. Mar 17, 2009

### 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).

9. Mar 17, 2009

### HallsofIvy

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]
and by adding $2\pi i$ repeatedly to $\theta$ you can get all n roots.

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.

(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.)

10. Mar 17, 2009