Proof of the pq formula: kzread.info/dash/bejne/gKCMwbiLXcieodI.html

Another way would be to just substitue x=a+bi directly where a and b are real and try to solve for a and b. (a+bi)^2 - 2i(a+bi) + 2 - 4i=0 a^2+2abi-b^2-2ai+2b+2-4i=0 [a^2-b^2+2b+2]+[2ab-2a-4]i=0+0i So a^2-b^2+2b+2=0 and 2ab-2a-4 =0 2ab-2a-4=0 ab-a-2=0 ab=a+2 b=1+2/a Substitute into the other equation. a^2-(1+2/a)^2+2(1+2/a)+2=0 a^2-(1+4/a+4/a^2)+2+4/a+2=0 a^2-1-4/a-4/a^2+2+4/a+2=0 a^2-4/a^2+3=0 a^4+3a^2-4=0 (a^2-1)(a^2+4)=0 a^2 = 1 or a^2 = -4(impossible since a is real) So a^2=1 a=1 or a=-1 b=1+2/1 b=1+2/(-1) =3 =-1 Thus the two answers are x=1+3i or x=-1-i This was actually the way i was taught to solve a quadratic equation if you get a complex number under the sqrt. It isn’t much easier but at least we get the answers immediately instead of needing to work out the sqrt first.

Thank you so much for sharing this. Regards 🙏

Another cool thing about squaring complex numbers. For the complex number: z = a + bi where you restrict a and b to be integers. z^2 in standard form will have real and imag. parts that will ALWAYS be the 2 smaller values (ignoring any - signs) of a Pythagorean Triplet! (2 - 3i)^2 = -5 - 12i . Now I know, strictly speaking, Pyth. Triplets a,b,c are always positive (for triangles) but you can extend them to the integers as a^2,b^2,c^2 are always positive i.e. (-5)^2 + (-12)^2 = (13)^2. It works every time. What this means for this video: In reverse, if you find the 2 square roots of a complex number that already has the 2 smaller values (ignoring the - sign) of a Pyth. Triplet for real and imag. parts you will ALWAYS get INTEGERS for the real and imag. parts of the 2 square roots. sqrt(-3+4i) = 1+3i or -1-i. Now, how cool is that! This was a "fudged" question as the quadratic equation was designed to have -3 and 4 in the discriminant. So who can be the first to offer a proof of this? It's not difficult - high school algebra is all that is needed.

@cheliu9140

2 ай бұрын

Assume that a, b are integers. We have: (a + bi)² = (a² - b²) + 2abi Set n = a² + b² We have: n = (a + bi)(a - bi) →n² = (a + bi)²(a - bi)² = [(a² - b²) + 2abi][(a² - b²) - 2abi] = (a² - b²)² + (2ab)² We found out that |a² - b²|, |2ab|, and n are parts of a Pythagorean triplet! (Credit: functor7)

@ianfowler9340

2 ай бұрын

@@cheliu9140 Good show!

You can always use the formula √(a+bi) = ½√(2√(a²+b²)+2a) + ½b̂i√(2√(a²+b²)−2a), where b̂ means 1 for b≥0 and −1 for b

What a sick time travel! That montage!

Great video as always! Just remember to say, "Real and Imaginary parts", as opposed to, "Real and Complex parts."

@david4649

2 ай бұрын

what's the difference?

@SalutLunar

2 ай бұрын

@@david4649 Complex numbers are encompassing of real and imaginary numbers. So all real numbers are complex numbers but not all complex numbers are real numbers. Likewise for imaginary numbers: all imaginary numbers are complex numbers but not all complex numbers are imaginary numbers. Have a look for blackpenredpen's video when he accidently labelled the axes on complex plane as Real and Complex. There were at least two videos apologising for/correcting the error.

There is an easier way to simplify sqrt(-3 + 4i). I realized -3 + 4i is a perfect square, so I did this: sqrt(-3 + 4i) = sqrt(1 + 4i - 4) = sqrt(1 + 2*2i + (2i)^2) = sqrt(1 + 2i)^2 = 1 + 2i (using the principal root mentioned in the video)

At 3:16 I would say -3+4i=5e^[i*atan2(4,-3)] and √(-3+4i=)=±√5e^[i*atan2(4,-3)/2]=±√5e(0.4472...+i*0.8944...)=±(1+2i)

How do I solve this to find x Sinx=1+Cosx

I feel like I've seen you do the "all over" mistake before. Is this a reupload, or did you almost make that same mistake twice when dealing with the pq formula?

@foogod4237

2 ай бұрын

It wouldn't be surprising to make that mistake more than once. The pq formula is actually just a simplified form of the quadratic formula, so when you're writing it out, if's not unreasonable for your brain to fall into that trap of trying to continue writing "the rest of it"..

8:18 ah yes exclude imaginary numbers where the entire video is an imaginary quadratic video

@bprpmathbasics

2 ай бұрын

a+bi, a and b have to be real.

@lizhang3073

2 ай бұрын

@@bprpmathbasics thanks for the clarification im not that good at math

Engrossing. Thanks.

I don't have any idea how to solve these equations but I can understand him solving it😮

asnwer=2g x

Compute the Integral x^2024/(x^2+1)^2023 dx

X²-x³=12, find value of x

There r only 2 comments?

@keescanalfp5143

2 ай бұрын

could mean that you have had the luck to not be the first .

I never learned about komplex numbers but i was curious… not anymore, i leave it to the mathematicians from this point 🫡🥲

@david4649

2 ай бұрын

Complex numbers really aren't that hard. I find it sad that they are not taught in school. A skill you need to have is knowing how to simplify, because without that skill, you won't be able to work with i. Schools make problems for calculators.