Solve 9+i/(2-2i)^2

$$\frac { 9 + i } { ( 2 - 2 i ) ^ { 2 } }$$

Evaluate

$$-\frac{1}{8}+\frac{9}{8}i=-0.125+1.125i$$

Solution Steps

Calculate 2-2i to the power of 2 and get -8i.

$$\frac{9+i}{-8i}$$

Multiply both numerator and denominator by imaginary unit i.

$$\frac{-1+9i}{8}$$

Divide -1+9i by 8 to get -\frac{1}{8}+\frac{9}{8}i.

$$-\frac{1}{8}+\frac{9}{8}i$$

Real Part

$$-\frac{1}{8} = -0.125$$

Solution Steps

Calculate 2-2i to the power of 2 and get -8i.

$$Re(\frac{9+i}{-8i})$$

Multiply both numerator and denominator of \frac{9+i}{-8i} by imaginary unit i.

$$Re(\frac{-1+9i}{8})$$

Divide -1+9i by 8 to get -\frac{1}{8}+\frac{9}{8}i.

$$Re(-\frac{1}{8}+\frac{9}{8}i)$$

The real part of -\frac{1}{8}+\frac{9}{8}i is -\frac{1}{8}.

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