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z=\frac{\left(1+3i\right)\left(2+i\right)}{\left(2-i\right)\left(2+i\right)}
Multiply both numerator and denominator of \frac{1+3i}{2-i} by the complex conjugate of the denominator, 2+i.
z=\frac{\left(1+3i\right)\left(2+i\right)}{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}.
z=\frac{\left(1+3i\right)\left(2+i\right)}{5}
By definition, i^{2} is -1. Calculate the denominator.
z=\frac{1\times 2+i+3i\times 2+3i^{2}}{5}
Multiply complex numbers 1+3i and 2+i like you multiply binomials.
z=\frac{1\times 2+i+3i\times 2+3\left(-1\right)}{5}
By definition, i^{2} is -1.
z=\frac{2+i+6i-3}{5}
Do the multiplications in 1\times 2+i+3i\times 2+3\left(-1\right).
z=\frac{2-3+\left(1+6\right)i}{5}
Combine the real and imaginary parts in 2+i+6i-3.
z=\frac{-1+7i}{5}
Do the additions in 2-3+\left(1+6\right)i.
z=-\frac{1}{5}+\frac{7}{5}i
Divide -1+7i by 5 to get -\frac{1}{5}+\frac{7}{5}i.