This problem was awesome! It is great to see you two collaborate!

@blackpenredpen

4 жыл бұрын

Eleazar Almazan thanks!!!

Next problem: ln(a+bi)

@benlev3375

4 жыл бұрын

Wouldn't that be Log(a+bi)?

@Ruslan-uv3xb

4 жыл бұрын

@@benlev3375 no,because log_10(z) doesn't equal log_e(z) or ln(z).

@Alians0108

4 жыл бұрын

@@Ruslan-uv3xb Log is used as based e usually.

@user-vs3lw6xs7n

4 жыл бұрын

@@Alians0108 no

@benlev3375

4 жыл бұрын

I studied complex numbers. Usually ln(z) has no meaning. Using Log(a+bi)=ln(abs(z))+iArg(z) with Arg(z) from -pi to pi is the correct notation.

It is easy as learning from two legends

@ProfOmarMath

4 жыл бұрын

Learning is a process that's done as a community, rather than a transfer of knowledge from one person to another 😍

Amazing formula! (The only thing I want to report: you switched the b0 at the end...😂 (not in the proof, but at 8:28))

You can use DeMoivre's thereom and the idea of how complex multiplication works to find any root of a complex number. Using the polar form would help alot.

@ProfOmarMath

4 жыл бұрын

Definitely. N-th roots even

actually the square root of (a) squared + (b) squared .. in the last formula just represent the modulus of a complex number

best explanation ive seen

thx man i needed just this

Impressive !

8:28 I think the conditions of b are reversed !

@blackpenredpen

4 жыл бұрын

Oh man, you are right. It's a typo. Thank you!!

We don't need to compute y, because yi has to be "root" conjugate regarding to x in case if dimension of x is equal to dimension of extended field. You can abstractly proof it from general polynomial equation. It corresponds to the group and field theory. And much more interesting is interpretation of the result, where perpendicular axis on rotational vector in its midpoint (somehow equal to rescaled its amplitude) creates two symmetric roots, such as we have in algorithm when solving quadratic equation by not so common way.

@ProfOmarMath

3 жыл бұрын

I personally think translating to polar form is the way to go on this one.

I think that tacitly choosing a branch of the logarithm can be misleading, and it is important not to sweep branch cuts under the rug. Once this has been done, you can define the square root of x as exp((1/2)*log(x)). This becomes increasingly important when dealing with integration, contours etc.

By the way Nice explanation 👍👍👍

De moivres theorem would help?

But why do u we not include a plus-minus sign before sqrt (a^2+b^2) in the two roots?

I wanted to try using a+bi = r×e^(i×thêta). We have sqrt(a+bi) = sqrt(r)×e^(i×thêta/2), so if we write sqrt(a+bi) as X+Yi, we get first that tan(thêta/2) = Y/X, and tan(thêta) = b/a Then we use tan(thêta) = tan(2×thêta/2) = 2×tan(thêta/2)/(1-tan(thêta/2)^2) = 2Y/(X(1-(Y^2/X^2))) = 2XY/(X^2 - Y^2) = b/a, which is basically the first equation b = 2XY, divided by the other one a = X^2 - Y^2. So there is a real number k so that 2XY = kb (1) X^2 - Y^2 = ka (2) We have also X^2 + Y^2 = sqrt(r)^2 = r = sqrt(a^2 + b^2) (3) (2)+(3) gives us that X^2 = (ka + sqrt(a^2 + b^2))/2 (2)-(3) gives that Y^2 = (-ka + sqrt(a^2 + b^2))/2 Then 2XY = sqrt( -(ka)^2 +ka×sqrt(...) -ka×sqrt(...) + a^2 + b^2) = sqrt(-(k^2 +1)×a^2 + b^2) So 4X^2 Y^2 = (-k^2 + 1)a^2 + b^2 = k^2 b^2 for any a,b so k = +-1 At this point I am close, but I can't find what is missing to find that k=1 without using identification of real and imaginary parts. If I have to resort to this identification method as he did, well it seems better (faster) to use his method because it leads to the same equations without introducing k. However, if there is another way to solve for k, then my calculus could be something else than just a longer path to the solution x) If somebody has an advice and the patience to read please comment. Sorry if there are mistakes especially about english.

@Vilinear

3 жыл бұрын

Hi, just tried this on my end. I wouldn't bother with all the 2XY stuff...once you have arctan(y/x) = theta, try drawing a triangle to deduce cos(theta) and sin(theta). The half-angle identities will give you the relevant results for cosine and sine of theta/2, from which the solution follows...

Personnally I would have transform the complex number under the form module, argument to work this out : the different signs will occur during the argument’s arctan phase according to the signs of a and b such as in your video.

@ProfOmarMath

4 жыл бұрын

This is a nice way to go. The variety of approaches is the spice of life 😍

👍

If b=0, then y=0 but not x : x=+/- sqrt(a) Nice video otherwise.

@blackpenredpen

4 жыл бұрын

If a

@egillandersson1780

4 жыл бұрын

@@blackpenredpen Sorry ! As I'm not english-speaker, I especially understand what is written 🤭

@blackpenredpen

4 жыл бұрын

Egill Andersson No need for sorry! : )

Next video about product of sum please

Name of the song?

@ProfOmarMath

4 жыл бұрын

I actually don't know but blackpenredpen might!

nice so such!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

This is a part of syllabus in our class 11 NCERT book and I just studied it a year ago .

@ProfOmarMath

3 жыл бұрын

Oh wow!

@akshaypanwar2393

3 жыл бұрын

@@ProfOmarMath Thank you sir 😊

@zerouno4477

3 жыл бұрын

That's why everybody recommend NCERT must important

@aashsyed1277

3 жыл бұрын

@@ProfOmarMath in pakistan & indiana EDIT: INDIA NOT INDIANA!

@mr_meow_77

2 жыл бұрын

@@aashsyed1277 Pakistan may bhi ncert chalte h ?

You should really get yourself a bigger whiteboard.

@ProfOmarMath

4 жыл бұрын

I think we should be kind to people and their resources and equipment during a pandemic and "stay at home" orders.

@WarpRulez

4 жыл бұрын

@@ProfOmarMath You make a good point.

@blackpenredpen

4 жыл бұрын

WarpRulez yea this is just temporary and I hope I can return to my classroom soon.

1:36 русские поймут лол

@fazex4185

4 жыл бұрын

?

@aashsyed1277

3 жыл бұрын

?

asnwer=(a+b)(a-b)=1 hello ha ri nang ha ri soon ha ri