How to solve for "a" | Oxford math admission test

[@boeubanks7507]

So, first of all, I dont believe that answer. The sum of the roots of a complex number can't equal a real number. Second, he made that problem way longer than it had to be. Pull the i out at the beginning, make root a common coefficient and pull it out front, multiply both sides by the complex conjugate, square both sides, and simplify. That is all you had to do to solve this. Simplicity is the hallmark of genius. Unnecessary complexity is the hallmark of a youtuber tying to make runtime.

[@paul2442]

I disagree, in fact theese roots of complex numbers add up to a real number. Another curious example of such coincidence is that i^i is a real number, even though at the first glance it seems that this has to be a (pure) complex number. But I agree that the solution in the video was unnecessary complicated.

[@meunknown5969]

I agree

[@adibbalkhiri9388]

kzbin.info/www/bejne/Zn6kZoGKoNx5nKssi=iw2sS4rDJcdoqT1E

[@EpsilonDeltaProof]

Nice, complex number

[@adibbalkhiri9388]

kzbin.info/www/bejne/Zn6kZoGKoNx5nKssi=iw2sS4rDJcdoqT1E

[@DominicMayers-zm3hv]

The square root is the solution with a positive real part, just as for the square root of a real number. sqrt -a makes an angle of 90 degrees with sqrt a in the complex plane, because they differ by the factor i. One must be at 45 degrees in the complex plane and the other at -45 degrees to allow the complex parts to cancel in the sum. The real part is 16, because they must add to 32. So sqrt a = 16 + 16 i = 16 (1 + i) and a = 16^2 (1+ i)^2 = 2 x 16^2 i. = 512 i. By symmetry, a = - 512 i also a solution.

[@adibbalkhiri9388]

kzbin.info/www/bejne/Zn6kZoGKoNx5nKssi=iw2sS4rDJcdoqT1E

[@tomaszjankowiak3777]

5 min without sens to have solutions possibile to reach in 10s.

[@adibbalkhiri9388]

kzbin.info/www/bejne/Zn6kZoGKoNx5nKssi=iw2sS4rDJcdoqT1E

[@NewsTopThe]

Nice solution.

[@MathBeast.channel-l9i]

Glad you think so! 😊

[@adibbalkhiri9388]

kzbin.info/www/bejne/Zn6kZoGKoNx5nKssi=iw2sS4rDJcdoqT1E

[@Maths_Se]

Very nice

[@RealQinnMalloryu4]

(16)+(16) =32 (2^3)+2^3) (2^1)+(2^1) (1)+(2)(a ➖ 2a+1)

[@adibbalkhiri9388]

kzbin.info/www/bejne/Zn6kZoGKoNx5nKssi=iw2sS4rDJcdoqT1E

[@nikko2505]

Можно было начать сразу с комплексных чисел. a^1/2(1+i)...

[@adibbalkhiri9388]

kzbin.info/www/bejne/Zn6kZoGKoNx5nKssi=iw2sS4rDJcdoqT1E