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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💜

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Comments

  1. PranavApril 29, 2021 at 3:31 PM

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

    ReplyDelete
  2. Amit April 29, 2021 at 6:10 PM

    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
  3. Best Digital Marketing CompanyFebruary 19, 2025 at 9:39 PM

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