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\frac{\left(2-2i\right)i}{-8i^{2}}
Multiply both numerator and denominator by imaginary unit i.
\frac{\left(2-2i\right)i}{8}
By definition, i^{2} is -1. Calculate the denominator.
\frac{2i-2i^{2}}{8}
Multiply 2-2i times i.
\frac{2i-2\left(-1\right)}{8}
By definition, i^{2} is -1.
\frac{2+2i}{8}
Do the multiplications in 2i-2\left(-1\right). Reorder the terms.
\frac{1}{4}+\frac{1}{4}i
Divide 2+2i by 8 to get \frac{1}{4}+\frac{1}{4}i.
Re(\frac{\left(2-2i\right)i}{-8i^{2}})
Multiply both numerator and denominator of \frac{2-2i}{-8i} by imaginary unit i.
Re(\frac{\left(2-2i\right)i}{8})
By definition, i^{2} is -1. Calculate the denominator.
Re(\frac{2i-2i^{2}}{8})
Multiply 2-2i times i.
Re(\frac{2i-2\left(-1\right)}{8})
By definition, i^{2} is -1.
Re(\frac{2+2i}{8})
Do the multiplications in 2i-2\left(-1\right). Reorder the terms.
Re(\frac{1}{4}+\frac{1}{4}i)
Divide 2+2i by 8 to get \frac{1}{4}+\frac{1}{4}i.
\frac{1}{4}
The real part of \frac{1}{4}+\frac{1}{4}i is \frac{1}{4}.