Complex Numbers Made EASY with Desmos!

[@webworkofficial]

but we still cant do complex number graph. like if i type abs(z)=4 it should make a circle of radius 4

[@AngieTeaches]

I know-I tried to do a complex graph when prepping this video, but no luck. 😞

[@dquixal77]

Given a complex number in its binomial form (a + bi) we have that its modulus is sqrt(a² + b²) as we already know from the Pythagorean theorem. For example, the number *2 + i* gives us *sqrt(2² + 1²)* = *sqrt(4 + 1)* = *sqrt(5)* . But thanks to Mr. Abraham De Moivre (1667 - 1754) we can transform this complex number into something else and eliminate this annoying square root and be left with only 5. De Moivre's theorem in _cis notation_ says the following: *(r * (cis theta))^n = r^n (cis n * theta)* That is to say, any complex number raised to the _n_ gives rise to the modulus raised to the _n_ of said complex number. The angle is of no concern to us. We will focus on the modulus exclusively. In our case _n_ equals 2. So we will square the complex number: *(a + bi)²* Unfolding Newton's Complex Binomial it is as follows: *a² - b² + 2abi* The module, when raised to the square, makes the square root disappear and we are left with *a² + b²* 2² - 1² + 2 * 2 * 1 * i = 4 - 1 + 4i = *3 + 4i* Starting from the complex numbers we have obtained a Pythagorean triplet!!! Modulus: *a² + b²* _(2² + 1² = 4 + 1 = 5)_ Leg a: *a² - b²* _(2² - 1² = 4 - 1 = 3)_ Leg b: *2ab* _(2 * 2 * 1 = 4)_ We can try with others: (3 + i)² = 8 + 6i = [6, 8, 10] (3 + 2i)² = 5 + 12i = [5, 12, 13] In Desmos: *z = a+bi* Sliders: *a = 2; b = 1* _(2 + i) we can try different values_ *m = b / a* _Slope_ *y=mx{0 < x < a}* _Modulus of z_ *w=a²-b²+2abi* _This is z²_ *y=(2ab/a²-b²)x{0 < x < a²-b²}* _Modulus of z²_

[@AngieTeaches]

Thanks for sharing this in-depth look at complex numbers!

[@dquixal77]

@@AngieTeaches I'm glad you like it. Working with complex numbers is something I wanted to do for a long time in Desmos. Thanks to your video, now I know I can do it. I had made a comment in Spanish for a video from another channel about Pythagorean triples and I want to share that with you. Now, it's me speaking, but I used Google Translate for the comment. I hope it's in proper english. "That is to say, any complex number raised to the n gives rise to the modulus raised to the n of said complex number" this part sounds weird to me. I completed the comment with some experimentation in Desmos. It was fun!