If a+ib = (root of 1+i) / (root of 1-i )then prove a^2 +b^2 =1
√(1+i)/√(1-i)
= √(1+i)√(1-i) / (1-i)
= √(1-i^2) / (1-i)
= √2/(1-i)
= √2(1+i) / (1-i)(1+i)
= √2(1+i)/2
= 1/√2 + 1/√2 i
a = 1/√2
b = 1/√2
a^2+b^2 = 1/2 + 1/2 = 1
or, using de Moivre's rule,
1+i = √2 cis π/4
so √(1+i) = ∜2 cis π/8
1-i = √2 cis -π/4
so √(1-i) = ∜2 cis -π/8
√(1+i)/√(1-i) = cis π/4
Now the angle doesn't matter, since cos^2+sin^2 = 1
To prove that a^2 + b^2 = 1, we first need to simplify the given expression.
Let's start by finding the square roots of 1+i and 1-i:
The number 1+i can be written in polar form as r1 * e^(iθ1), where r1 is the magnitude and θ1 is the angle.
For 1+i, the magnitude r1 is sqrt(1^2 + 1^2) = sqrt(2), and the angle θ1 is arctan(1/1) = π/4.
Similarly, for 1-i, the magnitude r2 is sqrt(1^2 + (-1)^2) = sqrt(2), and the angle θ2 is arctan((-1)/1) = -π/4.
Now, let's find the square roots using the expressions as r, r1, θ, and θ1:
√(1+i) = √(r * e^(iθ)) = √(r) * e^(iθ/2)
= √(sqrt(2)) * e^(i(π/4)/2)
= √(sqrt(2)) * e^(iπ/8)
= √(sqrt(2)) * (cos(π/8) + i*sin(π/8))
And, √(1-i) = √(r * e^(iθ)) = √(r) * e^(iθ/2)
= √(sqrt(2)) * e^(i(-π/4)/2)
= √(sqrt(2)) * e^(-iπ/8)
= √(sqrt(2)) * (cos(-π/8) + i*sin(-π/8))
= √(sqrt(2)) * (cos(π/8) - i*sin(π/8))
Now, substituting the values of √(1+i) and √(1-i) into the original expression:
a + ib = (√(sqrt(2)) * (cos(π/8) + i*sin(π/8))) / (√(sqrt(2)) * (cos(π/8) - i*sin(π/8)))
Both the numerator and denominator have the same magnitude, √(sqrt(2)), so they cancel out:
a + ib = (cos(π/8) + i*sin(π/8)) / (cos(π/8) - i*sin(π/8))
To simplify further, we can multiply the numerator and denominator by the conjugate of the denominator:
a + ib = [(cos(π/8) + i*sin(π/8)) * (cos(π/8) + i*sin(π/8))] / [(cos(π/8) - i*sin(π/8)) * (cos(π/8) + i*sin(π/8))]
Expanding the numerator and denominator:
a + ib = [(cos(π/8))^2 + 2i*cos(π/8)*sin(π/8) + (i*sin(π/8))^2] / [(cos(π/8))^2 - (i*sin(π/8))^2]
Note that (i^2) is equal to -1, so we can simplify further:
a + ib = [(cos(π/8))^2 + 2i*sin(π/8)*cos(π/8) - (sin(π/8))^2] / [(cos(π/8))^2 + (sin(π/8))^2]
Using the trigonometric identity sin^2(x) + cos^2(x) = 1:
a + ib = [(cos(π/8))^2 + 2i*sin(π/8)*cos(π/8) - (sin(π/8))^2] / 1
a + ib = [(cos(π/8))^2 - (sin(π/8))^2] + 2i*sin(π/8)*cos(π/8)
Applying the trigonometric identity cos(2x) = cos^2(x) - sin^2(x):
a + ib = cos(2*(π/8)) + i*sin(2*(π/8))
a + ib = cos(π/4) + i*sin(π/4)
Using the identity cos(π/4) = sin(π/4) = √(2)/2:
a + ib = √(2)/2 + i*√(2)/2
Comparing the real and imaginary parts:
a = √(2)/2
b = √(2)/2
Now, let's calculate a^2 + b^2:
a^2 + b^2 = (√(2)/2)^2 + (√(2)/2)^2
= 2/4 + 2/4
= 4/4
= 1
Therefore, we have proved that a^2 + b^2 = 1.