# In two equal complex numbers, what parts are equal to each other?

1. Jan 9, 2010

### Juwane

When there are, say, two complex numbers that are equal. What can we say about their equality? Can we say that the real part of one is equal to the real part of the other? Similarly, can we say that the complex part of one is equal to the complex part of the other?

Is this what it means when we say two complex numbers are equal?

2. Jan 9, 2010

yyes

3. Jan 9, 2010

### Juwane

Can we say this for complex numbers in polar form also? That is, can we say that their magnitudes and their angles are the same?

4. Jan 9, 2010

### nicksauce

Yes, as long as you have the constraint that $\theta\in[0,2\pi)$ (Or some other suitable interval).

5. Jan 9, 2010

### Juwane

But I don't see the intuition behind why, in two complex numbers, the real part of one is equal to only the real part of the other, and the same thing for the complex part.

6. Jan 9, 2010

### espen180

It is like a vector equation. You can think of a complex number a+bi as (a,b).

With vectors, [a,b]=[c,d] is equivalent to a=c and b=d. The same applies to complex numbers. (a,b)=(c,d) is equivalent to a=c and b=d.

It becomes clear when you visualize numbers in the complex plane as vectors from the origin to a point (a,b).

7. Jan 9, 2010

### fourier jr

or like any other constant c, x + cy = u + cv --> x=u & y=v & replace c with i

8. Jan 10, 2010

### HallsofIvy

Staff Emeritus
Because that is the way it is defined. "a+ bi= c+ di" is defined to mean "a= c and b= d". Surely that definition is in your text book.

9. Jan 10, 2010

Not only is it defined like that, but it makes sense intuitively in a geometric view. The two numbers are equal if they make the triangles they make with the axes are identical without transformation on an argand diagram.

The argand diagram view also shows an interpretation of why the components are separate, along the same reasons for vectors and vector spaces of any dimensionality. The components are independent... if you add a real number to a complex number, only the real component changes, and if you add an imaginary number to a complex number, only the imaginary component changes.

10. Jan 10, 2010

### Landau

This is plainly false. An extreme counter-example would be c=0, for then y and v can be anything. Complex numbers are (usually) defined as ordered pairs, and two ordered pairs (a,b) and (c,d) are defined to be (/characterized by being) equal iff a=c and b=d.

11. Jan 10, 2010

### payumooli

for a good answer refer to chapter 7
Introduction to Mathematical Philosophy
By Bertrand Russell

12. Jan 10, 2010

### Juwane

So let me get this straight.

If two complex numbers in rectangular form are equal to each other, then the real part and the imaginary part of one are equal to the corresponding parts of the other.

If two complex numbers in polar form are equal to each other, then the magnitude and the angle of one are equal to the corresponding parts of the other, except that there could be a multiple of 2pi in the angle of one of the numbers.

Am I right?

13. Jan 10, 2010

### wofsy

If they are equal they are equal - if you have a parameterization then their parameterizations must be equal - because they are equal.

14. Jan 11, 2010

### HallsofIvy

Staff Emeritus
Yes, that is exactly what has been said here and exactly what any book will give as the definition of the "rectangular" and "polar" forms for complex numbers.

The "$2\pi$" exception for polar form is because $\theta$ and $\theta+ 2\pi$ give exactly the same angle with the positive real axis.

15. Jan 11, 2010

### lugita15

Here's a fairly intuitive proof that equal complex numbers have equal real and imaginary parts.
Suppose that $$a+bi=c+di$$, but $$b \neq d$$. Then $$a-c=(d-b)i$$ and therefore $$i=\frac{a-c}{d-b}$$.
But this would mean that i is a real number, which is impossible. (In case you don't know why, square of a positive number is positive, the square of 0 is zero, and the square of a negative number is positive, no real number has a square which is negative, and hence i cannot be real).
So our original assumption is wrong, and thus equal complex numbers have equal imaginary parts, and $$b=d$$. But then $$a+bi=c+bi$$, which implies that $$a=c$$. So equal complex numbers have equal real parts as well.

Hope this helps.

16. Jan 12, 2010

### wofsy

Though all of the explanations here are correct I think they miss the point. A complex number is just a vector with two coordinates.

i is the vector (0,1). The purely imaginary axis is the y axis. These are the vectors (0,y) which by the definition of i is the same as iy. The purely real axis is the x axis. The vector (x,y) can be written as x + iy. This is nothing more than notation for the x,y coordinates of a 2 dimensional vector space.

Two vectors are equal iff their coordinates are equal. This has nothing to do with complex multiplication. It is just the equality of two vectors in a coordinate system.

Polar coordinates is just another coordinate system and works whether or not you think of the plane as a vector space or as the complex plane.

Last edited: Jan 12, 2010
17. Jan 12, 2010

### Landau

I think this point has been mentioned:

18. Jan 12, 2010

### wofsy

Right. I overlooked your post. Sorry.