If a+ib = (root of 1+i) / (root of 1-i )then prove a^2 +b^2 =1
To prove that a^2 + b^2 = 1, we need to simplify the expression a + ib = √(1 + i) / √(1 - i) into the form a^2 + b^2.
Let's start by finding the square roots of 1 + i and 1 - i:
√(1 + i):
To find the square root of 1 + i, we need to multiply the expression by its conjugate, which is 1 - i. This will help us in eliminating the imaginary term.
(1 + i) * (1 - i) / (1 - i) = (1 - i^2) / (1 - i) = (1 - (-1)) / (1 - i) = 2 / (1 - i)
√(1 + i) = √2 / (1 - i)
√(1 - i):
Following the same steps as above, we find:
√(1 - i) = √2 / (1 + i)
Now, let's substitute these values back into our original expression:
a + ib = (√2 / (1 - i)) / (√2 / (1 + i))
To divide by a fraction, we multiply by its reciprocal:
a + ib = (√2 / (1 - i)) * ((1 + i) / √2)
Simplifying, we get:
a + ib = (1 + i) / (1 - i)
Expanding the numerator and the denominator:
a + ib = (1 + i) * (1 + i) / (1 - i) * (1 + i)
Multiplying the numerator and denominator separately:
a + ib = (1 + 2i + i^2) / (1 - i + i - i^2)
Simplifying further:
a + ib = (1 + 2i - 1) / (1 + 1)
a + ib = (2i) / 2
Simplifying the expression:
a + ib = i
Comparing the real and imaginary parts on both sides of the equation, we have:
a = 0
b = 1
Now, let's substitute these values into a^2 + b^2:
a^2 + b^2 = 0^2 + 1^2
a^2 + b^2 = 0 + 1
a^2 + b^2 = 1
Therefore, we have proved that a^2 + b^2 = 1.
a+bi = √(1+i)/√(1-i)
= √(1+i)/√(1-i) * √(1-i)/√(1-i)
= √(1 - i^2) / (1-i)
= √(1 - i^2) / (1-i) * (1+i)/(1+i)
= √2(1+i)/2
= √2/2 + √2/2 i
so a = √2/2 and b=√2/2
and a^2 + b^2 = 2/4 + 2/4 = 1