Find the real and imaginary part of ( 1−𝑖 1+𝑖 ) 3

Question

Bytelearn · Accepted Answer

Simplify Expression: First, let's simplify the expression (1−𝑖)/(1+𝑖) before raising it to the power of 3. We can multiply the numerator and the denominator by the conjugate of the denominator to remove the imaginary unit from the denominator. The conjugate of (1+𝑖) is (1−𝑖). So, we multiply both the numerator and the denominator by (1−𝑖).Multiply by Conjugate: Perform the multiplication: [(1−𝑖)(1−𝑖)] / [(1+𝑖)(1−𝑖)] Now, expand both the numerator and the denominator. Numerator: (1−𝑖)(1−𝑖) = 1 - 2𝑖 + 𝑖^2 Denominator: (1+𝑖)(1−𝑖) = 1 - 𝑖^2Substitute i^2: Recall that 𝑖^2 = -1. Substitute 𝑖^2 with -1 in both the numerator and the denominator. Numerator: 1 - 2𝑖 - 1 = -2𝑖 Denominator: 1 - (-1) = 1 + 1 = 2 So, the simplified form is (-2𝑖) / 2.Divide and Simplify: Divide the numerator by the denominator: (-2𝑖) / 2 = -𝑖 Now we have the simplified form of the original expression, which is -𝑖.Raise to Power of 3: Raise the simplified expression to the power of 3: (-𝑖)^3 = (-𝑖)(-𝑖)(-𝑖)Calculate Power of -i: Calculate the power of -𝑖: (-𝑖)(-𝑖) = (-1)(𝑖)(-1)(𝑖) = (-1)(-1)(𝑖^2) = 1(-1) = -1 Now multiply by -𝑖: -1(-𝑖) = 𝑖Final Result: The final result is 𝑖, which means the real part is 0 and the imaginary part is 1.

Find the real and imaginary part of $\left(\frac{1-\mathit{i}}{1+\mathit{i}}\right)^3$

Full solution

More problems from Positive and negative square roots

Related topics

Find the real and imaginary part of (1−i1+i)3\left(\frac{1-\mathit{i}}{1+\mathit{i}}\right)^3(1+i1−i​)3

Find the real and imaginary part of $\left(\frac{1-\mathit{i}}{1+\mathit{i}}\right)^3$