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\frac{\left(2-5i\right)\left(3-2i\right)}{\left(3+2i\right)\left(3-2i\right)}
Multiply both numerator and denominator by the complex conjugate of the denominator, 3-2i.
\frac{\left(2-5i\right)\left(3-2i\right)}{3^{2}-2^{2}i^{2}}
Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.
\frac{\left(2-5i\right)\left(3-2i\right)}{13}
By definition, i^{2} is -1. Calculate the denominator.
\frac{2\times 3+2\times \left(-2i\right)-5i\times 3-5\left(-2\right)i^{2}}{13}
Multiply complex numbers 2-5i and 3-2i like you multiply binomials.
\frac{2\times 3+2\times \left(-2i\right)-5i\times 3-5\left(-2\right)\left(-1\right)}{13}
By definition, i^{2} is -1.
\frac{6-4i-15i-10}{13}
Do the multiplications in 2\times 3+2\times \left(-2i\right)-5i\times 3-5\left(-2\right)\left(-1\right).
\frac{6-10+\left(-4-15\right)i}{13}
Combine the real and imaginary parts in 6-4i-15i-10.
\frac{-4-19i}{13}
Do the additions in 6-10+\left(-4-15\right)i.
-\frac{4}{13}-\frac{19}{13}i
Divide -4-19i by 13 to get -\frac{4}{13}-\frac{19}{13}i.
Re(\frac{\left(2-5i\right)\left(3-2i\right)}{\left(3+2i\right)\left(3-2i\right)})
Multiply both numerator and denominator of \frac{2-5i}{3+2i} by the complex conjugate of the denominator, 3-2i.
Re(\frac{\left(2-5i\right)\left(3-2i\right)}{3^{2}-2^{2}i^{2}})
Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.
Re(\frac{\left(2-5i\right)\left(3-2i\right)}{13})
By definition, i^{2} is -1. Calculate the denominator.
Re(\frac{2\times 3+2\times \left(-2i\right)-5i\times 3-5\left(-2\right)i^{2}}{13})
Multiply complex numbers 2-5i and 3-2i like you multiply binomials.
Re(\frac{2\times 3+2\times \left(-2i\right)-5i\times 3-5\left(-2\right)\left(-1\right)}{13})
By definition, i^{2} is -1.
Re(\frac{6-4i-15i-10}{13})
Do the multiplications in 2\times 3+2\times \left(-2i\right)-5i\times 3-5\left(-2\right)\left(-1\right).
Re(\frac{6-10+\left(-4-15\right)i}{13})
Combine the real and imaginary parts in 6-4i-15i-10.
Re(\frac{-4-19i}{13})
Do the additions in 6-10+\left(-4-15\right)i.
Re(-\frac{4}{13}-\frac{19}{13}i)
Divide -4-19i by 13 to get -\frac{4}{13}-\frac{19}{13}i.
-\frac{4}{13}
The real part of -\frac{4}{13}-\frac{19}{13}i is -\frac{4}{13}.