Evaluate
-3+2i
Real Part
-3
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\frac{\left(2+3i\right)i}{-i^{2}}
Multiply both numerator and denominator by imaginary unit i.
\frac{\left(2+3i\right)i}{1}
By definition, i^{2} is -1. Calculate the denominator.
\frac{2i+3i^{2}}{1}
Multiply 2+3i times i.
\frac{2i+3\left(-1\right)}{1}
By definition, i^{2} is -1.
\frac{-3+2i}{1}
Do the multiplications in 2i+3\left(-1\right). Reorder the terms.
-3+2i
Divide -3+2i by 1 to get -3+2i.
Re(\frac{\left(2+3i\right)i}{-i^{2}})
Multiply both numerator and denominator of \frac{2+3i}{-i} by imaginary unit i.
Re(\frac{\left(2+3i\right)i}{1})
By definition, i^{2} is -1. Calculate the denominator.
Re(\frac{2i+3i^{2}}{1})
Multiply 2+3i times i.
Re(\frac{2i+3\left(-1\right)}{1})
By definition, i^{2} is -1.
Re(\frac{-3+2i}{1})
Do the multiplications in 2i+3\left(-1\right). Reorder the terms.
Re(-3+2i)
Divide -3+2i by 1 to get -3+2i.
-3
The real part of -3+2i is -3.
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