Solve (1+i)^7/1-i+(1-i)^7/1+i(3-4i)(2+2i)^3/4+3i

$$\frac { ( 1 + i ) ^ { 7 } } { 1 - i } + \frac { ( 1 - i ) ^ { 7 } } { 1 + i } \frac { ( 3 - 4 i ) ( 2 + 2 i ) ^ { 3 } } { 4 + 3 i }$$

Evaluate

$$136+128i$$

Solution Steps

Calculate 1+i to the power of 7 and get 8-8i.

$$\frac{8-8i}{1-i}+\frac{\left(1-i\right)^{7}}{1+i}\times \frac{\left(3-4i\right)\left(2+2i\right)^{3}}{4+3i}$$

Multiply both numerator and denominator of \frac{8-8i}{1-i} by the complex conjugate of the denominator, 1+i.

$$\frac{\left(8-8i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}+\frac{\left(1-i\right)^{7}}{1+i}\times \frac{\left(3-4i\right)\left(2+2i\right)^{3}}{4+3i}$$

Do the multiplications in \frac{\left(8-8i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}.

$$\frac{16}{2}+\frac{\left(1-i\right)^{7}}{1+i}\times \frac{\left(3-4i\right)\left(2+2i\right)^{3}}{4+3i}$$

Divide 16 by 2 to get 8.

$$8+\frac{\left(1-i\right)^{7}}{1+i}\times \frac{\left(3-4i\right)\left(2+2i\right)^{3}}{4+3i}$$

Calculate 1-i to the power of 7 and get 8+8i.

$$8+\frac{8+8i}{1+i}\times \frac{\left(3-4i\right)\left(2+2i\right)^{3}}{4+3i}$$

Multiply both numerator and denominator of \frac{8+8i}{1+i} by the complex conjugate of the denominator, 1-i.

$$8+\frac{\left(8+8i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}\times \frac{\left(3-4i\right)\left(2+2i\right)^{3}}{4+3i}$$

Do the multiplications in \frac{\left(8+8i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}.

$$8+\frac{16}{2}\times \frac{\left(3-4i\right)\left(2+2i\right)^{3}}{4+3i}$$

Divide 16 by 2 to get 8.

$$8+8\times \frac{\left(3-4i\right)\left(2+2i\right)^{3}}{4+3i}$$

Calculate 2+2i to the power of 3 and get -16+16i.

$$8+8\times \frac{\left(3-4i\right)\left(-16+16i\right)}{4+3i}$$

Multiply 3-4i and -16+16i to get 16+112i.

$$8+8\times \frac{16+112i}{4+3i}$$

Multiply both numerator and denominator of \frac{16+112i}{4+3i} by the complex conjugate of the denominator, 4-3i.

$$8+8\times \frac{\left(16+112i\right)\left(4-3i\right)}{\left(4+3i\right)\left(4-3i\right)}$$

Do the multiplications in \frac{\left(16+112i\right)\left(4-3i\right)}{\left(4+3i\right)\left(4-3i\right)}.

$$8+8\times \frac{400+400i}{25}$$

Divide 400+400i by 25 to get 16+16i.

$$8+8\left(16+16i\right)$$

Multiply 8 and 16+16i to get 128+128i.

$$8+\left(128+128i\right)$$

Add 8 and 128+128i to get 136+128i.

$$136+128i$$

Real Part

$$136$$

Solution Steps

Calculate 1+i to the power of 7 and get 8-8i.

$$Re(\frac{8-8i}{1-i}+\frac{\left(1-i\right)^{7}}{1+i}\times \frac{\left(3-4i\right)\left(2+2i\right)^{3}}{4+3i})$$

Multiply both numerator and denominator of \frac{8-8i}{1-i} by the complex conjugate of the denominator, 1+i.

$$Re(\frac{\left(8-8i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}+\frac{\left(1-i\right)^{7}}{1+i}\times \frac{\left(3-4i\right)\left(2+2i\right)^{3}}{4+3i})$$

Do the multiplications in \frac{\left(8-8i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}.

$$Re(\frac{16}{2}+\frac{\left(1-i\right)^{7}}{1+i}\times \frac{\left(3-4i\right)\left(2+2i\right)^{3}}{4+3i})$$

Divide 16 by 2 to get 8.

$$Re(8+\frac{\left(1-i\right)^{7}}{1+i}\times \frac{\left(3-4i\right)\left(2+2i\right)^{3}}{4+3i})$$

Calculate 1-i to the power of 7 and get 8+8i.

$$Re(8+\frac{8+8i}{1+i}\times \frac{\left(3-4i\right)\left(2+2i\right)^{3}}{4+3i})$$

Multiply both numerator and denominator of \frac{8+8i}{1+i} by the complex conjugate of the denominator, 1-i.

$$Re(8+\frac{\left(8+8i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}\times \frac{\left(3-4i\right)\left(2+2i\right)^{3}}{4+3i})$$

Do the multiplications in \frac{\left(8+8i\right)\left(1-i\right)}{\left(1+i\right)\left(1-i\right)}.

$$Re(8+\frac{16}{2}\times \frac{\left(3-4i\right)\left(2+2i\right)^{3}}{4+3i})$$

Divide 16 by 2 to get 8.

$$Re(8+8\times \frac{\left(3-4i\right)\left(2+2i\right)^{3}}{4+3i})$$

Calculate 2+2i to the power of 3 and get -16+16i.

$$Re(8+8\times \frac{\left(3-4i\right)\left(-16+16i\right)}{4+3i})$$

Multiply 3-4i and -16+16i to get 16+112i.

$$Re(8+8\times \frac{16+112i}{4+3i})$$

Multiply both numerator and denominator of \frac{16+112i}{4+3i} by the complex conjugate of the denominator, 4-3i.

$$Re(8+8\times \frac{\left(16+112i\right)\left(4-3i\right)}{\left(4+3i\right)\left(4-3i\right)})$$

Do the multiplications in \frac{\left(16+112i\right)\left(4-3i\right)}{\left(4+3i\right)\left(4-3i\right)}.

$$Re(8+8\times \frac{400+400i}{25})$$

Divide 400+400i by 25 to get 16+16i.

$$Re(8+8\left(16+16i\right))$$

Multiply 8 and 16+16i to get 128+128i.

$$Re(8+\left(128+128i\right))$$

Add 8 and 128+128i to get 136+128i.

$$Re(136+128i)$$

The real part of 136+128i is 136.

$$136$$