Question

If \[ (x+iy)=\left(\frac{1+i}{1-i}\right)^3-\left(\frac{1-i}{1+i}\right)^3, \] then the true statement among the following is:

• A. \(x\lt y\)
• B. \(x\gt y\)
• C. \(x\neq 0\)
• D. \(x=y\)

Hint:

For complex number simplification, rationalize the denominator first and then use powers of \(i\): \[ i^2=-1,\quad i^3=-i,\quad i^4=1 \] to simplify expressions quickly.

Answer 1

The Correct Option is B

Step 1: Simplify \[ \frac{1+i}{1-i} \] Multiply numerator and denominator by \[ 1+i \] Then, \[ \frac{1+i}{1-i}\times \frac{1+i}{1+i} = \frac{(1+i)^2}{1-i^2} \] Since \[ i^2=-1, \] we get \[ 1-i^2=1+1=2 \] Also, \[ (1+i)^2=1+2i+i^2 \] \[ =1+2i-1 \] \[ =2i \] Hence, \[ \frac{1+i}{1-i}=\frac{2i}{2}=i \] Therefore, \[ \left(\frac{1+i}{1-i}\right)^3=i^3=-i \]

Step 2: Simplify \[ \frac{1-i}{1+i} \] Multiply numerator and denominator by \[ 1-i \] Then, \[ \frac{1-i}{1+i}\times \frac{1-i}{1-i} = \frac{(1-i)^2}{1-i^2} \] Again, \[ 1-i^2=2 \] Now, \[ (1-i)^2=1-2i+i^2 \] \[ =1-2i-1 \] \[ =-2i \] Hence, \[ \frac{1-i}{1+i}=\frac{-2i}{2}=-i \] Therefore, \[ \left(\frac{1-i}{1+i}\right)^3=(-i)^3=i \]

Step 3: Find \(x+iy\).
Substituting the values, \[ x+iy=(-i)-i \] \[ x+iy=-2i \] Comparing real and imaginary parts, \[ x=0 \] and \[ y=-2 \]

Step 4: Compare \(x\) and \(y\).
We have \[ 0\gt -2 \] Thus, \[ x\gt y \]

Step 5: Final conclusion.
Therefore, \[ \boxed{x\gt y} \]