Solve (1+i)^4/(1-i)^3+(1-i)^4/(1+i)^3

$$\frac { ( 1 + i ) ^ { 4 } } { ( 1 - i ) ^ { 3 } } + \frac { ( 1 - i ) ^ { 4 } } { ( 1 + i ) ^ { 3 } }$$

Evaluate

$$2$$

Solution Steps

Calculate 1+i to the power of 4 and get -4.

$$\frac{-4}{\left(1-i\right)^{3}}+\frac{\left(1-i\right)^{4}}{\left(1+i\right)^{3}}$$

Calculate 1-i to the power of 3 and get -2-2i.

$$\frac{-4}{-2-2i}+\frac{\left(1-i\right)^{4}}{\left(1+i\right)^{3}}$$

Multiply both numerator and denominator of \frac{-4}{-2-2i} by the complex conjugate of the denominator, -2+2i.

$$\frac{-4\left(-2+2i\right)}{\left(-2-2i\right)\left(-2+2i\right)}+\frac{\left(1-i\right)^{4}}{\left(1+i\right)^{3}}$$

Do the multiplications in \frac{-4\left(-2+2i\right)}{\left(-2-2i\right)\left(-2+2i\right)}.

$$\frac{8-8i}{8}+\frac{\left(1-i\right)^{4}}{\left(1+i\right)^{3}}$$

Divide 8-8i by 8 to get 1-i.

$$1-i+\frac{\left(1-i\right)^{4}}{\left(1+i\right)^{3}}$$

Calculate 1-i to the power of 4 and get -4.

$$1-i+\frac{-4}{\left(1+i\right)^{3}}$$

Calculate 1+i to the power of 3 and get -2+2i.

$$1-i+\frac{-4}{-2+2i}$$

Multiply both numerator and denominator of \frac{-4}{-2+2i} by the complex conjugate of the denominator, -2-2i.

$$1-i+\frac{-4\left(-2-2i\right)}{\left(-2+2i\right)\left(-2-2i\right)}$$

Do the multiplications in \frac{-4\left(-2-2i\right)}{\left(-2+2i\right)\left(-2-2i\right)}.

$$1-i+\frac{8+8i}{8}$$

Divide 8+8i by 8 to get 1+i.

$$1-i+\left(1+i\right)$$

Add 1-i and 1+i to get 2.

$$2$$

Real Part

$$2$$

Solution Steps

Calculate 1+i to the power of 4 and get -4.

$$Re(\frac{-4}{\left(1-i\right)^{3}}+\frac{\left(1-i\right)^{4}}{\left(1+i\right)^{3}})$$

Calculate 1-i to the power of 3 and get -2-2i.

$$Re(\frac{-4}{-2-2i}+\frac{\left(1-i\right)^{4}}{\left(1+i\right)^{3}})$$

Multiply both numerator and denominator of \frac{-4}{-2-2i} by the complex conjugate of the denominator, -2+2i.

$$Re(\frac{-4\left(-2+2i\right)}{\left(-2-2i\right)\left(-2+2i\right)}+\frac{\left(1-i\right)^{4}}{\left(1+i\right)^{3}})$$

Do the multiplications in \frac{-4\left(-2+2i\right)}{\left(-2-2i\right)\left(-2+2i\right)}.

$$Re(\frac{8-8i}{8}+\frac{\left(1-i\right)^{4}}{\left(1+i\right)^{3}})$$

Divide 8-8i by 8 to get 1-i.

$$Re(1-i+\frac{\left(1-i\right)^{4}}{\left(1+i\right)^{3}})$$

Calculate 1-i to the power of 4 and get -4.

$$Re(1-i+\frac{-4}{\left(1+i\right)^{3}})$$

Calculate 1+i to the power of 3 and get -2+2i.

$$Re(1-i+\frac{-4}{-2+2i})$$

Multiply both numerator and denominator of \frac{-4}{-2+2i} by the complex conjugate of the denominator, -2-2i.

$$Re(1-i+\frac{-4\left(-2-2i\right)}{\left(-2+2i\right)\left(-2-2i\right)})$$

Do the multiplications in \frac{-4\left(-2-2i\right)}{\left(-2+2i\right)\left(-2-2i\right)}.

$$Re(1-i+\frac{8+8i}{8})$$

Divide 8+8i by 8 to get 1+i.

$$Re(1-i+\left(1+i\right))$$

Add 1-i and 1+i to get 2.

$$Re(2)$$

The real part of 2 is 2.

$$2$$