Nice. You might enjoy looking up "polar form for a complex number." I think you'll be pleasantly surprised at how powerful it is for doing certain calculations! z = re^(i*theta) !!
Yes, to multiply complex numbers just add the angles and multiply the magnitudes. 'i' has an angle of 90 (or pi/2) and magnitude of 1 so the solution has an angle of 45 or 225 and a magnitude of 1. Much simpler and you get to draw pretty pictures with x's and o's
@@thisnamewastakentoo_ “Simpler” isn’t necessarily “better”. I have seen far too many math profs teach the “simplest” approach, which is usually unnecessarily abstract and often so slick as to seem mystifying. The approach is this video is great! If the YTer wants to learn more, well, I pointed him in the right direction.
@@jeremyalm9006 I was following up from your comment that polar coordinates are powerful and giving an example of how easy this problem is when using them. Simpler is better when you use it. Relying on what you know is better when you learn it. Complex math links together so many branches of math from exponents to trig and geometry allowing them to work together to solve problems. I have still not forgiven my math teachers for making us learn a bunch of trig identities when knowing basic exponent rules and the Euler formula covers everything.
That's a great question but, no it would just be those two answers shown in the video. Let's start with a simpler explanation first, the square root of 9 is 3 which can be rewritten as (1+2). We can multiply that entire thing by -1 which gives up (-1-2) which equals -3 which is the other square root of 9. This means we have to multiply the entire thing by -1 because both (-1+2) = 1 and (1-2) = -1 are not square roots of 9. Let's look at the problem now. If we use sqrt(2)/2 - (sqrt(2)/2)i or (sqrt(2)/2)i - sqrt(2)/2 and squared both of them it would result in -i. Let us look at why this is with (sqrt(2)/2 - (sqrt(2)/2)i)^2. We need to foil so let's write it out like (sqrt(2)/2 - (sqrt(2)/2)i) * (sqrt(2)/2 - (sqrt(2)/2)i). Let's foil it and we get (1/2) - (1/2)i -(1/2)i - (1/2). the 1/2 cancel out which just leaves -i.
So √𝒊 = ±(1 + 𝒊)/√2, right? Both values. How do we then calculate their sum √𝒊 + √𝒊 or product (√𝒊)·(√𝒊) Which of the two values we take for each of the roots to evaluate the result?
And what is a principal branch anyway? As far as I know, it may (if there is really a need for it) be defined in various ways, depending on what we think the more appropriate choice is in the context of the problem we are solving. There are at least *_two_* pretty "standard" ways to define a principal branch that you may come across in math literature, and the "principal" values of *√(−𝒊)* for them would be different.
after a^2 - b^2 = 0, 2ab = 1, couldn't you use equation 1 to say a^2 = b^2, therefore a = b, then use 2a^2 = 4, therefore a = ±√2/2 = b? if a = -b, then plugging back in and squaring doesn't work, so it's an extraneous solution (√2/2 - i√2/2)^2 = 1/2 - 1/2 - 2(1/2)i = -i.
That is a great observation that I didn't realize until I was re-watching the video yesterday but yes you could do that with equation 1 and it would be much easier. 👍
Very clear and helpful explanation. Thanks!
Are the Oompa Loompas learning algebra?
Now that was fun!
❤❤❤
Cool video
Thanks mate!
Should i call it "root of imagination"?
Thats not an offical name or anything but I just found that a fun thing to name it.
You can do it in complex polar coordinate in your head.
Nice. You might enjoy looking up "polar form for a complex number." I think you'll be pleasantly surprised at how powerful it is for doing certain calculations! z = re^(i*theta) !!
Yes, to multiply complex numbers just add the angles and multiply the magnitudes. 'i' has an angle of 90 (or pi/2) and magnitude of 1 so the solution has an angle of 45 or 225 and a magnitude of 1. Much simpler and you get to draw pretty pictures with x's and o's
@@thisnamewastakentoo_ “Simpler” isn’t necessarily “better”. I have seen far too many math profs teach the “simplest” approach, which is usually unnecessarily abstract and often so slick as to seem mystifying.
The approach is this video is great! If the YTer wants to learn more, well, I pointed him in the right direction.
@@jeremyalm9006 I was following up from your comment that polar coordinates are powerful and giving an example of how easy this problem is when using them. Simpler is better when you use it. Relying on what you know is better when you learn it. Complex math links together so many branches of math from exponents to trig and geometry allowing them to work together to solve problems. I have still not forgiven my math teachers for making us learn a bunch of trig identities when knowing basic exponent rules and the Euler formula covers everything.
@@thisnamewastakentoo_ Fair enough.
In the second step, when u square both sides, don't u square them independently, and if not then why
woul there alsobe other solutions having a mix of the -ve and +ve solutions for a and b? like sqrt(2)/2 - (sqrt(2)/2) i or (sqrt(2)/2 )i - sqrt(2)/2?
That's a great question but, no it would just be those two answers shown in the video. Let's start with a simpler explanation first, the square root of 9 is 3 which can be rewritten as (1+2). We can multiply that entire thing by -1 which gives up (-1-2) which equals -3 which is the other square root of 9. This means we have to multiply the entire thing by -1 because both (-1+2) = 1 and (1-2) = -1 are not square roots of 9.
Let's look at the problem now. If we use sqrt(2)/2 - (sqrt(2)/2)i or (sqrt(2)/2)i - sqrt(2)/2 and squared both of them it would result in -i. Let us look at why this is with (sqrt(2)/2 - (sqrt(2)/2)i)^2. We need to foil so let's write it out like (sqrt(2)/2 - (sqrt(2)/2)i) * (sqrt(2)/2 - (sqrt(2)/2)i). Let's foil it and we get (1/2) - (1/2)i -(1/2)i - (1/2). the 1/2 cancel out which just leaves -i.
@@NoahBugbee Ohhh i see now thank уou!
So √𝒊 = ±(1 + 𝒊)/√2, right?
Both values.
How do we then calculate their sum
√𝒊 + √𝒊
or product
(√𝒊)·(√𝒊)
Which of the two values we take for each of the roots to evaluate the result?
Probably positive due to principle branch
But in the video it was shown that there are *_two_* complex values of the square root.
And what is a principal branch anyway? As far as I know, it may (if there is really a need for it) be defined in various ways, depending on what we think the more appropriate choice is in the context of the problem we are solving. There are at least *_two_* pretty "standard" ways to define a principal branch that you may come across in math literature, and the "principal" values of *√(−𝒊)* for them would be different.
I have a less controversial name AN ORTHOGONAL NUMBER
Bro next video on de moivre's theorem
I will certainly make a vid on it soon.👍
De Moivre’s anyone?
Planning to make a video on it very soon.👍
My calculator gives up on this. I press '=' and it does nothing.
Most likely your calculator does not do complex numbers.
-1
after a^2 - b^2 = 0, 2ab = 1, couldn't you use equation 1 to say a^2 = b^2, therefore a = b, then use 2a^2 = 4, therefore a = ±√2/2 = b? if a = -b, then plugging back in and squaring doesn't work, so it's an extraneous solution (√2/2 - i√2/2)^2 = 1/2 - 1/2 - 2(1/2)i = -i.
That is a great observation that I didn't realize until I was re-watching the video yesterday but yes you could do that with equation 1 and it would be much easier. 👍