Solve 5i/2-i+5i

$$\frac { 5 i } { 2 - i } + 5 i$$

Evaluate

$$-1+7i$$

Solution Steps

Multiply both numerator and denominator of \frac{5i}{2-i} by the complex conjugate of the denominator, 2+i.

$$\frac{5i\left(2+i\right)}{\left(2-i\right)\left(2+i\right)}+5i$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{5i\left(2+i\right)}{2^{2}-i^{2}}+5i$$

By definition, i^{2} is -1. Calculate the denominator.

$$\frac{5i\left(2+i\right)}{5}+5i$$

Multiply 5i times 2+i.

$$\frac{5i\times 2+5i^{2}}{5}+5i$$

By definition, i^{2} is -1.

$$\frac{5i\times 2+5\left(-1\right)}{5}+5i$$

Do the multiplications in 5i\times 2+5\left(-1\right). Reorder the terms.

$$\frac{-5+10i}{5}+5i$$

Divide -5+10i by 5 to get -1+2i.

$$-1+2i+5i$$

Combine the real and imaginary parts in numbers -1+2i and 5i.

$$-1+\left(2+5\right)i$$

Add 2 to 5.

$$-1+7i$$

Real Part

$$-1$$

Solution Steps

Multiply both numerator and denominator of \frac{5i}{2-i} by the complex conjugate of the denominator, 2+i.

$$Re(\frac{5i\left(2+i\right)}{\left(2-i\right)\left(2+i\right)}+5i)$$

Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$Re(\frac{5i\left(2+i\right)}{2^{2}-i^{2}}+5i)$$

By definition, i^{2} is -1. Calculate the denominator.

$$Re(\frac{5i\left(2+i\right)}{5}+5i)$$

Multiply 5i times 2+i.

$$Re(\frac{5i\times 2+5i^{2}}{5}+5i)$$

By definition, i^{2} is -1.

$$Re(\frac{5i\times 2+5\left(-1\right)}{5}+5i)$$

Do the multiplications in 5i\times 2+5\left(-1\right). Reorder the terms.

$$Re(\frac{-5+10i}{5}+5i)$$

Divide -5+10i by 5 to get -1+2i.

$$Re(-1+2i+5i)$$

Combine the real and imaginary parts in numbers -1+2i and 5i.

$$Re(-1+\left(2+5\right)i)$$

Add 2 to 5.

$$Re(-1+7i)$$

The real part of -1+7i is -1.

$$-1$$