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z=\frac{5i\left(2+i\right)}{\left(2-i\right)\left(2+i\right)}+5i
Multiply both numerator and denominator of \frac{5i}{2-i} by the complex conjugate of the denominator, 2+i.
z=\frac{5i\left(2+i\right)}{2^{2}-i^{2}}+5i
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{5i\left(2+i\right)}{5}+5i
By definition, i^{2} is -1. Calculate the denominator.
z=\frac{5i\times 2+5i^{2}}{5}+5i
Multiply 5i times 2+i.
z=\frac{5i\times 2+5\left(-1\right)}{5}+5i
By definition, i^{2} is -1.
z=\frac{-5+10i}{5}+5i
Do the multiplications in 5i\times 2+5\left(-1\right). Reorder the terms.
z=-1+2i+5i
Divide -5+10i by 5 to get -1+2i.
z=-1+\left(2+5\right)i
Combine the real and imaginary parts in numbers -1+2i and 5i.
z=-1+7i
Add 2 to 5.