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]?
>>7812176
wait why doesn't the latex work
>>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]
You can't split roots unless the inner terms are multiplying or dividing.
>>7812180
Use spaces in your LaTeX or 4chan will add them for you.
Test:
[math]2 + 2 = topkek[[/math]
Test: [math]2 + 2 = topkek[[/math]
>>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?
>>7812202
No.
Because then it wouldn't be a real number dipshit.
>>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.
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]?
>>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.
>>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).
>>7812223
Sorry this is a better link:
https://en.wikipedia.org/wiki/Euler%27s_formula
>>7812202
no because the reals are a subset of the complex.
>>7812162
Funny how it does make sense if you square both sides and take the real part.
>>7812202
Top laff
What does [math] 0.5 + 0.i [/math] mean?
>>7812589
It means 0.5.
>>7812609
And the 0.i? Just a nice way of saying no imaginary part?
>>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
>>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.