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Is this even possible?
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You are currently reading a thread in /sci/ - Science & Math

Thread replies: 25
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650 B, 188x20
Is this even possible?
[math]\sqrt{6} = \sqrt{9-3} = 3+i\sqrt{3}[/math]
>>
Are you implying that [math]\sqrt{x}+\sqrt{y}=\sqrt{x+y}[/math]?
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>>7812176
wait why doesn't the latex work
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>>7812162
No, you can't break up sums in a radical.

For example, [math]16=\sqrt{256}=\sqrt{64+64+64+64} \neq \sqrt{64}+\sqrt{64}+\sqrt{64}+\sqrt{64}=8*4=32[/math]
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>>7812187
[eqn] 16 = \sqrt{256} = \sqrt{64+64+64+64} \neq \sqrt{64} + \sqrt{64} + \sqrt{64} + \sqrt{64} = 8 * 4 = 32 [/eqn]

>>7812180
It breaks sometimes for me too.
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You can't split roots unless the inner terms are multiplying or dividing.
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>>7812180
Use spaces in your LaTeX or 4chan will add them for you.
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>>7812190
here
>>7812180
I added spaces in between everything. Also used [eqn] tags but the spaces might have been enough.
>>
Test:
[math]2 + 2 = topkek[[/math]

Test: [math]2 + 2 = topkek[[/math]
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>>7812190
Holy shit I'm really fucking stupid.
Wait so are there any way to write a real number as a complex number with a non-zero immaginary part?
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>>7812202
No.

Because then it wouldn't be a real number dipshit.
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>>7812202
Not really.

A real number [math]r[/math] can be written as [math]r + 0i[/math].
Assume it can also be written as [math]a + bi[/math], with [math]a , b[/math] real and [math]b \neq 0[/math].
Then, [math]r + 0i = a + bi[/math] implies [math]\frac{r-a}{b} = i[/math]. But this doesn't make any sense, as the left-hand side is a real number while the right side isn't.
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One more question: the modulus of a complex number written in trigonometric form is the same when the complex number is written in imaginary exponential?
[math] z e^{ i\alpha } = z \cos \alpha + i\sin \alpha [/math]?
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>>7812202
No, that contradicts the definition of a real number.

You sound like you'd be interested in laplace transforms though. They might be a bit above your head but if you're genuinely interested I'd look into it.
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>>7812217
That's Euler's equation.
https://en.wikipedia.org/wiki/Euler%27s_identity
However the z should multiply the entire thing (both the cosine and the sine).
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>>7812223
Sorry this is a better link:
https://en.wikipedia.org/wiki/Euler%27s_formula
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>>7812223
Yeah i forgot the parenthesis. I've never seen euler's formula with a number before [math]e[/math] so I assumed it couldn't have a modulus.

>>7812219
Thanks, I know I won't understand everything but I like reading wikipedia article regarding math, just to grasp the concept.
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>>7812202
no because the reals are a subset of the complex.
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>>7812162
Funny how it does make sense if you square both sides and take the real part.
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>>7812202
Top laff
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What does [math] 0.5 + 0.i [/math] mean?
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>>7812589
It means 0.5.
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>>7812609
And the 0.i? Just a nice way of saying no imaginary part?
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>>7812217
You're missing brackets before cos and ending after alpha, but yeah regardless of how you express a complex number always has the same modulus
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>>7812621
Wherever you found the expression "0.5 + 0.i" was generated by some software that someone wrote and forgot to account for the special case where the imaginary part is exactly 0.

0.i means as much as saying 0.x.
Thread replies: 25
Thread images: 1
Thread DB ID: 468858



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