If α denotes the number of solutions of |1-i|ˣ=2ˣ and β=(|z|/(arg (z))), where z=π/4(1+i)⁴[(1-√π i)/(√π+i)+(√π-i)/(1+√π i)], i=√-1, then the distance of the point (α, β) from the line 4 x-3 y=7 is __________.

Solution

Maths sir · 4 steps

Spot that each denominator is just i times its numerator, and the messy bracket collapses to -2i.

1. Find alpha. |1-i| = √2, so the equation is 2^(x/2) = 2ˣ. That forces x/2 = x, so x = 0 is the only solution and α = 1.
2. Simplify the bracket. i(1-√π i) = √π + i and i(√π - i) = 1 + √π i. So each fraction equals 1/i = -i, and the bracket is -2i.

   z = π/4(1+i)⁴(-2i) = π/4(-4)(-2i) = 2π i

3. Find beta. (1+i)² = 2i, so (1+i)⁴ = (2i)² = -4. Then z = 2π i lies on the positive imaginary axis: |z| = 2π and arg z = π/2, so β = 2π/(π/2) = 4.
4. So. The distance of (1, 4) from 4x - 3y - 7 = 0 is (|4 - 12 - 7|)/√(16+9) = 15/5 = 3.

The common slip: 1.8. Taking (1+i)⁴ = 4 instead of -4 gives z = -2π i, arg z = -π/2, β = -4 and a distance of 1.8.

The idea to keep. When a fraction looks like (a+bi)/(b-ai), check whether the denominator is ± i times the numerator. One multiplication by i often kills the whole mess.

Answer from the NTA final answer key. Solution by PrepWiser.