Solve (3-i)i^3/1-2i

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

Evaluate

$$1-i$$

Solution Steps

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

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

Multiply 3-i and -i to get -1-3i.

$$\frac{-1-3i}{1-2i}$$

Multiply both numerator and denominator by the complex conjugate of the denominator, 1+2i.

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

Do the multiplications in \frac{\left(-1-3i\right)\left(1+2i\right)}{\left(1-2i\right)\left(1+2i\right)}.

$$\frac{5-5i}{5}$$

Divide 5-5i by 5 to get 1-i.

$$1-i$$

Real Part

$$1$$

Solution Steps

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

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

Multiply 3-i and -i to get -1-3i.

$$Re(\frac{-1-3i}{1-2i})$$

Multiply both numerator and denominator of \frac{-1-3i}{1-2i} by the complex conjugate of the denominator, 1+2i.

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

Do the multiplications in \frac{\left(-1-3i\right)\left(1+2i\right)}{\left(1-2i\right)\left(1+2i\right)}.

$$Re(\frac{5-5i}{5})$$

Divide 5-5i by 5 to get 1-i.

$$Re(1-i)$$

The real part of 1-i is 1.

$$1$$