Solve -7i/2+5i

$$\frac { - 7 i } { 2 + 5 i }$$

Evaluate

$$-\frac{35}{29}-\frac{14}{29}i\approx -1.206896552-0.482758621i$$

Solution Steps

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

$$\frac{-7i\left(2-5i\right)}{\left(2+5i\right)\left(2-5i\right)}$$

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

$$\frac{-7i\left(2-5i\right)}{2^{2}-5^{2}i^{2}}$$

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

$$\frac{-7i\left(2-5i\right)}{29}$$

Multiply -7i times 2-5i.

$$\frac{-7i\times 2-7\left(-5\right)i^{2}}{29}$$

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

$$\frac{-7i\times 2-7\left(-5\right)\left(-1\right)}{29}$$

Do the multiplications in -7i\times 2-7\left(-5\right)\left(-1\right). Reorder the terms.

$$\frac{-35-14i}{29}$$

Divide -35-14i by 29 to get -\frac{35}{29}-\frac{14}{29}i.

$$-\frac{35}{29}-\frac{14}{29}i$$

Real Part

$$-\frac{35}{29} = -1\frac{6}{29} = -1.206896551724138$$

Solution Steps

Multiply both numerator and denominator of \frac{-7i}{2+5i} by the complex conjugate of the denominator, 2-5i.

$$Re(\frac{-7i\left(2-5i\right)}{\left(2+5i\right)\left(2-5i\right)})$$

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

$$Re(\frac{-7i\left(2-5i\right)}{2^{2}-5^{2}i^{2}})$$

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

$$Re(\frac{-7i\left(2-5i\right)}{29})$$

Multiply -7i times 2-5i.

$$Re(\frac{-7i\times 2-7\left(-5\right)i^{2}}{29})$$

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

$$Re(\frac{-7i\times 2-7\left(-5\right)\left(-1\right)}{29})$$

Do the multiplications in -7i\times 2-7\left(-5\right)\left(-1\right). Reorder the terms.

$$Re(\frac{-35-14i}{29})$$

Divide -35-14i by 29 to get -\frac{35}{29}-\frac{14}{29}i.

$$Re(-\frac{35}{29}-\frac{14}{29}i)$$

The real part of -\frac{35}{29}-\frac{14}{29}i is -\frac{35}{29}.

$$-\frac{35}{29}$$