Solve 1/i-3i/1-i

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

Evaluate

$$\frac{3}{2}-\frac{5}{2}i=1.5-2.5i$$

Solution Steps

Multiply both numerator and denominator of \frac{1}{i} by imaginary unit i.

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

Divide i by -1 to get -i.

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

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

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

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

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

Divide -3+3i by 2 to get -\frac{3}{2}+\frac{3}{2}i.

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

Add -i and \frac{3}{2}-\frac{3}{2}i to get \frac{3}{2}-\frac{5}{2}i.

$$\frac{3}{2}-\frac{5}{2}i$$

Real Part

$$\frac{3}{2} = 1\frac{1}{2} = 1.5$$

Solution Steps

Multiply both numerator and denominator of \frac{1}{i} by imaginary unit i.

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

Divide i by -1 to get -i.

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

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

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

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

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

Divide -3+3i by 2 to get -\frac{3}{2}+\frac{3}{2}i.

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

Add -i and \frac{3}{2}-\frac{3}{2}i to get \frac{3}{2}-\frac{5}{2}i.

$$Re(\frac{3}{2}-\frac{5}{2}i)$$

The real part of \frac{3}{2}-\frac{5}{2}i is \frac{3}{2}.

$$\frac{3}{2}$$