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\frac{\left(10+8i\right)i}{-9i^{2}}
Multiply both numerator and denominator by imaginary unit i.
\frac{\left(10+8i\right)i}{9}
By definition, i^{2} is -1. Calculate the denominator.
\frac{10i+8i^{2}}{9}
Multiply 10+8i times i.
\frac{10i+8\left(-1\right)}{9}
By definition, i^{2} is -1.
\frac{-8+10i}{9}
Do the multiplications in 10i+8\left(-1\right). Reorder the terms.
-\frac{8}{9}+\frac{10}{9}i
Divide -8+10i by 9 to get -\frac{8}{9}+\frac{10}{9}i.
Re(\frac{\left(10+8i\right)i}{-9i^{2}})
Multiply both numerator and denominator of \frac{10+8i}{-9i} by imaginary unit i.
Re(\frac{\left(10+8i\right)i}{9})
By definition, i^{2} is -1. Calculate the denominator.
Re(\frac{10i+8i^{2}}{9})
Multiply 10+8i times i.
Re(\frac{10i+8\left(-1\right)}{9})
By definition, i^{2} is -1.
Re(\frac{-8+10i}{9})
Do the multiplications in 10i+8\left(-1\right). Reorder the terms.
Re(-\frac{8}{9}+\frac{10}{9}i)
Divide -8+10i by 9 to get -\frac{8}{9}+\frac{10}{9}i.
-\frac{8}{9}
The real part of -\frac{8}{9}+\frac{10}{9}i is -\frac{8}{9}.