Let m=±ai where a>0 is real. Then m and -m lie on opposite sides of a radius a circle centered on the origin of the complex plane. Now √(±m) = √a√(±i)=√a[cos(±π/4)+i*sin(±π/4)] Then √m+√(-m)=√a[cos(π/4)+i*sin(π/4)]+√a[cos(-π/4)+i*sin(±π/4)]=2√a(√2/2)=12 It follows that √a=12/√2, a=144/2=72, and m=±72i.
Am a chemist but after following you , am now more mathematician than a chemist.
Correct
√m +√-m = 12 -> √m(1+i) = 12 -> √m = 12/(1+i) -> m = ±12²/(1+i)² -> m = ±144/(1-1+2i) -> m=±72/i = ∓72i (Fastest Method )
Arbitrary approach:
Given equation is
√m + √( - m) = 12 ...(1)
Let √m = p and √( - m) = q
Then m = p² and - m = q²
Then p² + q² = 0 ...(2)
Also equation (1) becomes
p + q = 12 ...(3)
Now, (p + q)² = p² + 2pq + q²
= (p² + q²) + 2pq
Putting p + q = 12 and p² + q² = 0, we get
(12)² = 0 + 2pq
or, 144 = 2pq
or, 72 = pq ...(4)
Now, (p - q)² = (p + q)² - 4pq
= (12)² - 4(72)
= 144 - 288
= - 144
= 144i²
= (12i)²
=> p - q = ± 12i
Now, 2p = (p + q) + (p - q)
= 12 + ( ± 12i )
= 12 + 12i, 12 - 12i
=> p = 6 + 6i, 6 - 6i
=> √m = 6 + 6i, 6 - 6i
=> m = (6 + 6i)², (6 - 6i)²
= 6² + 2(6)(6i) + (6i)²,
6² - 2(6)(6i) + (6i)²
= 36 + 72i + 36i²,
36 - 72i + 36i²
= 36 + 72i + ( - 36 ) , 36 - 72i + ( - 36 )
= 72i, - 72i
= ± 72i
Another approach :
Let √m = p and √( - m) = q
Then p + q = 12 ...(1)
Also, m = p² and - m = q²
Now, p² - q² = m - ( - m)
or, p² - q² = 2m
or, (p + q)(p - q) = 2m
or, 12 (p - q) = 2m
or, p - q = m/6
Now, 2p = (p + q) + (p - q)
or, 2p = 12 + (m/6)
or, 2√m = 12 + (m/6)
or, (2√m)² = { 12 + (m/6) }²
or, 4m = (12)² + 2(12)(m/6) + (m/6)²
or, 4m = 144 + (24)(m/6) + (m²/36)
or, 4m = 144 + 4m + (m²/36)
or, 0 = 144 + (m²/36)
or, 144 + (m²/36) = 0
or, (m²/36) = - 144
or, m² = - 36 × 144
or, m² = - 6² × 12²
or, m² = - (6 × 12)²
or, m² = - 72²
or, m² = 72²i²
or, m² = (72i)²
or, m = ± 72i
And many more approaches...
Let m=±ai where a>0 is real. Then m and -m lie on opposite sides of a radius a circle centered on the origin of the complex plane.
Now √(±m) = √a√(±i)=√a[cos(±π/4)+i*sin(±π/4)] Then √m+√(-m)=√a[cos(π/4)+i*sin(π/4)]+√a[cos(-π/4)+i*sin(±π/4)]=2√a(√2/2)=12
It follows that √a=12/√2, a=144/2=72, and m=±72i.
Sqrt[72i]+Sqrt[-72i] m=±72i It’s in my head.
Nice one, that is while you are the best sir.
Maximum respect sir.