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Let's complex bash Part 1

I have to learn complex bash. And almost everyone knows that I am notes taking girl so thought why not make a post on complex bash ( so that I don't get emotionally demotivated lol).😇 There wasn't any need for learning complex bash, but it was in my dream checklist i.e " To learn a bash." And since I am not loaded with exams, I think it's high time to learn Bash and new topics. 

Also if anyone from the "anti-bash" community is reading, sorry in advance and R.I.P. 



Notes:-

1. Complex numbers are of the form z=a+ib, where a and b are real numbers and i^2=-1.

2. In polar form, z=r(\cos \theta+~~i\sin\theta)=~~re^{i\theta}, where r=~~|z|=~~\sqrt{a^2+b^2}, which is called the magnitude.

3. Here we used euler's formula i.e \cos \theta+~~i\sin\theta=~~e^{i\theta}.

4. The \theta is called the argument of z, denored \arg z. ( \theta can be considered in \mod 360 and it is  measured anti-clockwise).

5. The complex conjugate of z is denoted as \bar{z}.

6. We have \bar{z}=~~a-bi=~~re^{-i\theta}. It is the reflection of the point wrt the real axis.

7. The properties:- 

  • \overline{p+q}=\bar{p}+\bar{q}
  • \overline{p-q}=\bar{p}-\bar{q}
  • \overline{p\cdot q}=\bar{p}\cdot \bar{q}
  • \overline{p/q}=\bar{p}/\bar{q}
8. Verrification:- Let p=a+bi and q=c+di.

Then we have p+q=~~(a+c)+~~(b+d)i. 

So \boxed{\overline{p+q}=~~(a+c)-~~(b+d)i=~~ a-bi+~~c-di=~~\bar{p}+\bar{q}}. Similarly for \overline{p-q}. 

Now, for \overline{p\cdot q}. We have p\cdot q=~~(a+bi)\cdot(c+di)=~~ac-bd+~~(ad+bc)i..

So \boxed{\overline{p\cdot q}=~~(a+bi)\cdot(c+di) =~~ac-bd-~~(ad+bc)i=~~(a-bi)\cdot(c-di)}. Similarly for \overline{p/q}.

9. If we have z=a+ib, then we have a^2+b^2=~~|z|^2=~~(a+bi)(a-bi)=~~z\bar{z}.

See you in next post 

Sunaina💜

Comments

  1. Umm, this is the part of complex numbers, where is bashing here 🤔

    ReplyDelete
  2. If you are learning complex numbers you can try problem 9 from this month's Ganit Bikash.I solved it using complex roots of polynomials.

    ReplyDelete

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