Complex numbers : Finding Square roots of : ExamSolutions

[@ExamSolutions_Maths]

Because i equals the root of -1 and so when you square this you get -1... -1 multiplied by b squared gives - b squared. Take a look at my early videos on complex numbers to see these basic results for squaring, cubing etc for values of i.

[@peterjhw07]

Thank you that was really helpful. Excellently explained, thanks again.

[@halimabegum4630]

I'm not sure. I just wanted an explanation on how to factorise those quartic equations that you did in this video

[@AliShaikh_744]

Use u substitution. Substitute b^2 for u and it becomes a quadratic. Solve for u. Once solved for u, replace u again with b^2 and then solve for b.

[@emmydee408]

@@AliShaikh_744 Thank you brother 😊❤️, I really appreciate it, I now understand

[@AliShaikh_744]

@@emmydee408 Glad I was able to help😁

[@SM-ly5tf]

How can I pay you money?

[@masterbaits4108]

you can now donate to him! please consider if possible though it's completely understandable if you cant due to the pandemic :)

[@ExamSolutions_Maths]

Thanks

[@MORIANWANJIKU]

God bless you

[@halimabegum4630]

Do you have a video up on how to factorise quartic equations?

[@avalonase6356]

Thank you man may god bless u😊

[@trkingdavid4943]

i like the way u interpret the question

[@izhakalmakhadi5563]

Thanks mate

[@ExamSolutions_Maths]

No problem

[@haidermansoor378]

you are a saviour 🙏🙏🙏

[@BossDropbear]

(a+bi)^2 =(a^2-b^2)+2abi 7-24i So .. ab=-12 and (a-b)(a+b)=7 a=-4, b=3 is a solution, as is a=4, b=-3. i.e. -4+3i, 4-3i.

[@leetong2628]

why solution for square root of y must be real number ? 6:35 and 11:30

[@junaidwali8938]

If b isn't real and has an i component, the final answer would be a+bi^2. Which is equal to a-b, and there would be no imaginery parts in the final answer, thus it would not be a complex number[

[@radheypatel9622]

@@junaidwali8938 you're about 3 years and 6 months late

[@modreplayy2]

... time travel@@radheypatel9622

[@BlackAngel-mf6do]

​@@radheypatel9622 better late than never😅

[@lsmith3942]

why does b have to be a real value 6:40?

[@bostangpalaguna228]

It will yield the same result by considering only b real. The a and b will only be swapped

[@saferehman8234]

If b is not real then it would make a+bi into just a+b (as i x i is real) and so it won't make it correct statement

[@folabi1380]

Safe Rehman ah ok

[@chewphuiyee]

Sir, may I know at the beginning part, how did you make that ...+b squared i squared reduced to -b squared?

[@aguyontheinternet8436]

I know it's been 9 years, but seriously? That was the most difficult part to understand?

[@grassytramtracks]

It's because it's (bi)² so b²i² i²=-1 so it's b²×-1 so -b²

[@bilbobaggins9321]

There are 2 types of people lmao

[@ExamSolutions_Maths]

Do you mean ones with complex roots?

[@ExamSolutions_Maths]

b is a real number. For example 2i is an imaginary number but the 2 is real

[@Womiwomiwomiaaahahhh]

i needed this

[@farshiddianat8068]

but if you substitute B= -5, +5 into the first equation (a^2-b^2= -21) you would only get a=2 right?

[@QuickOffline]

Thanks a lot

[@chewphuiyee]

Thank you very much!

[@SoftwareSith]

Quality video. Thanks for the help.

[@ExamSolutions_Maths]

Cheers

[@ExamSolutions_Maths]

+faiza anwar At 6.44 I said that b was a real number. It is not an imaginary number. 3i is imaginary but the 3 is real. So bi is imaginary but b is a real number. I hope that makes sense.

[@dataenthusiast6721]

Very helpful.

[@prachinpatel7242]

You sir are a legend thank you so much

[@ExamSolutions_Maths]

Best wishes - thanks for your support.

[@modreplayy2]

... if I may ask... why isn't b = +/- 2i one of the solutions ?

[@coolsigmaalpha]

because in this case, B is the co-efficient of the imaginary number and +/- 2 was gained by solving for the real number

[@heishillary]

Can't believe this was uploaded when i was 10 years old

[@hamsaudaykiran8373]

Super

[@wdfomfg]

e^ipi = -1, i^2ipi = 1, 4(e^2ipi) = 4

[@APOSTASA]

if i want to find the square root of 1+root of 2 for example can i use a similar method?

[@ExamSolutions_Maths]

Yes

[@dhvsheabdh]

Sorry, my brother logged into my e-mail account and replied to this. Rest assured, I understood it. Well played sir.

[@gamuchirainmago3328]

nice

[@dope_axel9436]

😭😭😭you saved me

[@dhvsheabdh]

-(i^2) st.

[@wdfomfg]

Oh okay, and now i noticed i made a typo in the second statement :O that i should be an e....

[@joycew4586]

but if you use b=12/a, the answer is ±(4+3i) instead of ±(4-3i) does it matter? :) thanks

[@ExamSolutions_Maths]

b = -12/a not 12/a

[@NdabenhlePhakathi]

How is that a negative B squared

[@bournelucid]

Why did the sign not change when the 21b^2 switched sides

[@cowboyengineer9227]

did you ever figure this out haha

[@grassytramtracks]

Because it didn't switch sides, he added b⁴ and and subtracted 100 to make it 0=b⁴-21b²-100 but just wrote it the other way round as b⁴-21b²-100

[@danieloh1353]

Sugene Kim approved

[@adilalinadir9739]

what if there was a "z" in the complex equation ?

[@saferehman8234]

Z is not part of a complex equation z IS a complex equation

[@recipeworld4339]

gud

[@dhvsheabdh]

Haha it seems so! I didn't bother to look, I read the first comment anyway.

[@dhvsheabdh]

*first ;)

[@dhvsheabdh]

???

[@cianstapleton7238]

f

[@kalyanv3092]

Or we can use polar forms

[@yess.6558]

Thank you!