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Imaginary numbers


blackhole123

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have you tried MathWorld? Wikipedia or any other encyclopedia?

 

But, seriously, I have my doubts that there could ever be a "one place" where you could have everything. I mean, just on this one topic of imaginary numbers (complex analysis), there are several multi-hundred page books listed as still being in print on Amazon.com. Add to that all the journal articles and out of print texts on complex analysis, and we're talking hundreds of thousands of pages. And, all that on just one small part of mathematics. Collecting it all together is probably an impossible task.

 

I've heard of these nearly magical rooms which are filled with thousands of books and quiet people... I'm not sure I believe these myths though ;)

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I've heard of these nearly magical rooms which are filled with thousands of books and quiet people... I'm not sure I believe these myths though ;)

 

Ah, but no library has "everything" and even then, the information about complex analysis would be spread around in many different books and journals. It isn't all collected into one single spot. It's an impossible task, I tell ya.

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so a complex number with the real part as 2 and the imaginary part as 3 is written as

[math]n = 2 + 3i[/math]

 

What does the 4 represent in [math]n^2=4+2*2...[/math]?

 

That's [math](a+b)^2 =a^2 + 2ab + b^2[/math] so [math] (2+3i)^2 = 2^2 + 2 * (2*3i) + (3i)^2[/math]. The 4 is the [math]2^2[/math].

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(2,3) is not a complex number. It is a point in R2. [math]n=2+3i[/math] is a complex number, and [math]n^2=4+2*2*3i+(3i)^2=-5+12i[/math]

 

[math]\mathbb{C} \cong \mathbb{R}^{2}[/math]

 

So all alan2here was asking was can you multiply complex numbers (essentially). Yes, of course as shown by the earlier posts.

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Actually I was looking for the method also. TY for the equations.

 

Edit:

Oh, how can

[math]n^2=4+2*2*3i+(3i)^2=-5+12i[/math]

 

You're using add, times and power on positive numbers. How can a negative number be created for the real part?

Edited by alan2here
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Foil, looks useful.

 

Where N is a complex number

the real part of [math]n^2=4+r*r*(i*-3)[/math]

Is this right, and what is the equation for the imaginary part?

 

No. The formula for the square of a binomial expression is [math](a+b)^2 = a^2 + 2ab + b^2[/math]. If you're working with an imaginary number [math]n^2=(a,b)^2=(a+bi)^2=a^2+2\* a\* b\* i + (bi)^2[/math]. The new real part would be [math]a^2+ (bi)^2=a^2-b^2[/math] and the imaginary part would be [math] 2\*a\*b\*i[/math].

 

Your main mistake is that the real part is only factored in once, it is an unfortunate that the example you gave had [math]a=2[/math] so [math] 2\*a\*b[/math] became ambiguous. Your second mistake seems to be a lapse in understanding of the relationship between [math](a,b)[/math] and [math](a+bi)[/math] as well as what FOIL (poduct of the First two numbers plus product of the Outside two numbers plus product of the Inside two numbers plus product of the Last two numbers) and how it simplifies to the square of a binomial formula due to the identity [math](a+b)^2 = (a+b)\*(a+b)[/math].

Edited by Kyrisch
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Thanks

 

Please note that the following is case sensitive

 

So to put it another way

[math]n[/math] = complex number

 

[math]i[/math] = imaginary part of [math]n[/math]

[math]r[/math] = real part of [math]n[/math]

 

 

[math]n = n ^ 2[/math]

 

To do this I must

 

[math]R = r ^ 2 - i ^ 2[/math]

 

[math]I = 2 * r * i[/math]

 

[math]r = R[/math]

 

[math]i = I[/math]

 

 

And it works :¬)

Edited by alan2here
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