# Questions about imaginary number and root of 4

1. Apr 5, 2008

### Shing

I am thinking that
if the imaginary number is bigger than the other number,
is it right to say that:
i> 5 ?
7i> 3i ?
Does i has magnitude?

if
$$Z_1=4+5i$$
then
$$Z_2=1-3i$$
whether $$Z_1>Z_2 or Z_2>Z_1$$ is true?

If we say $$Z_a$$ is bigger than $$Z_b$$, does that means the absolute value of these complex number?

Thank you :)

Last edited: Apr 5, 2008
2. Apr 5, 2008

### Shing

Also
I have been wonder why $$\sqrt{4} =2$$
but why NOT -2, negative two?

3. Apr 5, 2008

### arildno

No, "i" is not bigger than any real number.
As it happens, when we talk about complex and imaginary numbers, we cannot have an ordering relation between them.
Numerous complex numbers will have the same magnitude, so that a number's magnitude cannot serve as an identificatory trait of that number.

4. Apr 5, 2008

### arildno

Good question!

Because we have, by convenience, DEFINED the square root thusly.

Similarly, by convenience, we have defined 1 not to be a prime number.

5. Apr 5, 2008

### HallsofIvy

Staff Emeritus
It is not possible to assign an "order" to the complex numbers in such a way as to have an "ordered field" (that is, so that if a< b and 0< c, then ac< bc and if a< b, then a+c< b+ c (for any c)). For example, if we were to define an order so that 0< i, then we must have 0*i< i*i or 0< -1. Since this is not necessarily "regular order" that is not a contradiction itself but multiplying by i again, 0*i< -1*i or 0< -i. Adding i to both sides, we must have 0+i< -i+ i or i< 0, contradicting 0< i. But if we try to define an order so that i< 0, we can, in the same way, show that 0< i getting the same contradiction.

Neither is true- as I just showed there is no way to compare complex numbers.

I've never seen anyone say that Za> Zb for Za and Zb complex numbers. If you mean to say one has larger absolute value than the other, then you must say |Za|> |Zb|.

We define it that way because fractional powers with even denominators give real results only for positive numbers. If we want to be able to say that $\sqrt{x}= x^{1/2}$ (which is a very useful thing to do) and then combine it with other such functions, we need to stick to positive numbers.

6. Apr 7, 2008

### Shing

Thank you so much, Arildno and HallsofIvy!!

But I don't understand that,would you explain a bit more please?
imaginary number is so amusing, I am thinking of the geometry meaning of some operations of complex number.

I was thinking if these are true:
$$Z_1\times Z_2$$ produce a new complex number "vector" $$Z_3$$
$${Z_1}^{1/2}\times {Z_2}^{1/2}$$produce a new complex number$$Z_3$$ too

When I was thinking of the reciprocal of a complex number $$Z_1$$

$${1\over {Z_1}} = {{x-iy}\over {x^2+y^2}}$$

I was shocked!!:surprised

Does that mean $${1\over {Z_1}}$$ is $$Z_1$$* times the reciprocal of the area of the square magnitude of $$Z_1$$($$|Z_1|^2$$)?

If so, what is the meaning of a reciprocal of an area?

Also, is $${1\over {Z_1}}$$ still a complex number?

7. Apr 7, 2008

### Pere Callahan

Yes, it is, unless Z_1 is zero.

Try to relate the geometric meaning of 1/z to the circle inversion.

8. Apr 7, 2008

### nandu11

"i" is nothin..as we say an imaginary number n nothin else..when we say that Za is greater than Zb..we mean to take their real part n not the imaginary parts

9. Apr 7, 2008

### Diffy

You can however compare absolute values of complexes. If you have $$Z_1=4+5i$$ and $$Z_2=1-3i$$ then you can look at $$|Z_1|$$ and $$|Z_2|$$.

This will tell you the complex numbers distance away from zero in the complex plane.

In general let $$Z = a+Bi$$ then $$|Z| = \sqrt{a^2 + b^2}$$

Now you can say $$|Z_1|>|Z_2|$$.

10. Apr 7, 2008

### HallsofIvy

Staff Emeritus
I have never seen such a statement! And, exactly what do you mean by '"i" is nothin"?

11. Apr 7, 2008

### HallsofIvy

Staff Emeritus
I have never seen anyone say "Za> Zb" when they meant Re(Z_a)> Re(Z_b). And what do you mean by ' "i" is nothin'?