Solve 5+2i/2-5i

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

Evaluate

$$i$$

Solution Steps

Multiply both numerator and denominator by the complex conjugate of the denominator, 2+5i.

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

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

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

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

$$\frac{\left(5+2i\right)\left(2+5i\right)}{29}$$

Multiply complex numbers 5+2i and 2+5i like you multiply binomials.

$$\frac{5\times 2+5\times \left(5i\right)+2i\times 2+2\times 5i^{2}}{29}$$

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

$$\frac{5\times 2+5\times \left(5i\right)+2i\times 2+2\times 5\left(-1\right)}{29}$$

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

$$\frac{10+25i+4i-10}{29}$$

Combine the real and imaginary parts in 10+25i+4i-10.

$$\frac{10-10+\left(25+4\right)i}{29}$$

Do the additions in 10-10+\left(25+4\right)i.

$$\frac{29i}{29}$$

Divide 29i by 29 to get i.

$$i$$

Real Part

$$0$$

Solution Steps

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

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

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

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

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

$$Re(\frac{\left(5+2i\right)\left(2+5i\right)}{29})$$

Multiply complex numbers 5+2i and 2+5i like you multiply binomials.

$$Re(\frac{5\times 2+5\times \left(5i\right)+2i\times 2+2\times 5i^{2}}{29})$$

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

$$Re(\frac{5\times 2+5\times \left(5i\right)+2i\times 2+2\times 5\left(-1\right)}{29})$$

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

$$Re(\frac{10+25i+4i-10}{29})$$

Combine the real and imaginary parts in 10+25i+4i-10.

$$Re(\frac{10-10+\left(25+4\right)i}{29})$$

Do the additions in 10-10+\left(25+4\right)i.

$$Re(\frac{29i}{29})$$

Divide 29i by 29 to get i.

$$Re(i)$$

The real part of i is 0.

$$0$$