How can I prove that if wz = 0, then w = 0 or z = 0 in complex analysis?

[Physics_wiz]

Prove that if wz = 0, then w = 0 or z = 0. w and z are two complex numbers.
I said that w = a + bi and z = c + di and set wz = 0. I got down to c(a+b) = d(b-a), but don't know where to go from here.

I'm trying to teach myself complex analysis, anyone know any good sources? I took the 3 calculus classes and an elementary differential equations class. I'm a mechanical engineering major though and I don't think I'll have time to take a complex analysis class unless I want to stay in college after I get my BS in ME.

 

[NateTG]

You can solve it like a system of variables:
[tex]w=a+bi[/itex]
[tex]z=c+di[/itex]
[tex]wz=0[/itex]
$$wz=(a+bi)(c+di)=(ac-bd)+(ad+bc)i$$
so
$$ac-bd=0$$
and
[tex]ad+bc=0[/itex]
so
[tex]d=\frac{-bc}{a}[/itex]
so
$$ac+\frac{b^2c}{a}=0$$
$$a^2c+b^2c=0$$
so (case i)
$$a^2+b^2=0$$
since $a$ and $b$ are real numbers this gives that $z=0$
otherwise (case ii) c=0
then
[tex]bd=0[/itex]
and
[tex]ad=[/itex]
so, $d=0$ so we have $w=0$ or $a=0 \rm{and} b=0$ and we have $z=0$.

 

[Physics_wiz]

Got it. Thanks.
Know any good sources to learn complex analysis from?

EDIT: I guess it would be smart to know how big of a problem I'm trying to tackle here. Is self-teaching complex analysis going to be a hard task? Is it comparable to...say, someone who only knows algebra trying to teach himself calculus?

 

[hypermorphism]

I like the approach used in Tristan Needham's Visual Complex Analysis, but as you've not yet built up any intuition for complex numbers, you may want to accompany it with a more traditional text.

 

[matt grime]

Surely it is far easier to simply use |vw|=|v||w|? This isn't complex analysis.

 

[Physics_wiz]

More questions

I attached a file with three more questions in it. I did the first two using:
z = x + yi and w = a + bi and solved them down in the same manner like the first question I posted. Is there an easier way to do these?

I'm not really sure what to do for #6 though. Do I just set z = x + yi and find an equation f(y) = f(x) then graph it on an x-y plane? I did that for part (a) and got a circle with center (2, -3) and radius 2, but is that what I'm supposed to do?

Thanks.

 

[hypermorphism]

Try thinking about what the equations are saying geometrically. In the case of the first one that you solved, it is saying that the distance between the points z and the point (2,-3) is a constant 2. That's just a circle of radius 2 centered at (2,-3) with no algebra necessary.
The second equation says that the distance between the points z and the point (0,2) is less than or equal to 1, which describes a closed disc around that point of radius 1. What about the geometry described by (d) ?
Use the algebraic method only if the geometry is hopelessly obscured by the equation. Even then, you can now use the geometric result you found algebraically for later questions.

 

[Physics_wiz]

Ahh thanks a lot...that makes things a bit easier. See if I got those right:

b) The distance between z and -2i is less than or equal to 1. Which means it's a disk centered at -2i with a radius of 1.
c) The real part of z is equal to 4. Straight vertical line at 4 because the imaginary part can take on any value.
d) The distance between z and (1, -2i) is equal to the distance between z and (-3, -i). z again can be any value on a straight line that's a perpendicular bisector to the line connecting (1, -2i) and (-3, -i).
e) Ellipse with focii (-1, 0i) and (1, 0i).
f) Hyperbola with the same focii as part e.

By the way, when you write (2, -3) are you talking about the point a = 2 - 3i in the complex plane? When I graph those lines and ellipses and circles, do I graph them all in a complex plane with an imaginary axis and a real one?

 

[hypermorphism]

Yep, all those interpretations are spot on.
If you take the complex plane to be a 2-dimensional real vector space with basis vectors 1 and i, then you can write the element a+bi as the (position) vector (a,b) and preserve mathematical rigour.