Solve -1+2i/2+i

$$\frac { - 1 + 2 i } { 2 + i }$$

Evaluate

$$i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 2-i.

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

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

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

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

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

Multiply complex numbers -1+2i and 2-i like you multiply binomials.

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

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

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

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

$$\frac{-2+i+4i+2}{5}$$

Combine the real and imaginary parts in -2+i+4i+2.

$$\frac{-2+2+\left(1+4\right)i}{5}$$

Do the additions in -2+2+\left(1+4\right)i.

$$\frac{5i}{5}$$

Divide 5i by 5 to get i.

$$i$$

Real Part

$$0$$

Solution Steps

Multiply both numerator and denominator of \frac{-1+2i}{2+i} by the complex conjugate of the denominator, 2-i.

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

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

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

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

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

Multiply complex numbers -1+2i and 2-i like you multiply binomials.

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

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

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

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

$$Re(\frac{-2+i+4i+2}{5})$$

Combine the real and imaginary parts in -2+i+4i+2.

$$Re(\frac{-2+2+\left(1+4\right)i}{5})$$

Do the additions in -2+2+\left(1+4\right)i.

$$Re(\frac{5i}{5})$$

Divide 5i by 5 to get i.

$$Re(i)$$

The real part of i is 0.

$$0$$