Solve 2-3i/4+5i-2+5i^2

$$\frac{ 2-3i }{ 4+5i } -2+5 { i }^{ 2 }$$

Evaluate

$$-\frac{294}{41}-\frac{22}{41}i\approx -7.170731707-0.536585366i$$

Solution Steps

Multiply both numerator and denominator of \frac{2-3i}{4+5i} by the complex conjugate of the denominator, 4-5i.

$$\frac{\left(2-3i\right)\left(4-5i\right)}{\left(4+5i\right)\left(4-5i\right)}-2+5i^{2}$$

Do the multiplications in \frac{\left(2-3i\right)\left(4-5i\right)}{\left(4+5i\right)\left(4-5i\right)}.

$$\frac{-7-22i}{41}-2+5i^{2}$$

Divide -7-22i by 41 to get -\frac{7}{41}-\frac{22}{41}i.

$$-\frac{7}{41}-\frac{22}{41}i-2+5i^{2}$$

Do the additions.

$$5i^{2}-\frac{89}{41}-\frac{22}{41}i$$

Calculate i to the power of 2 and get -1.

$$5\left(-1\right)-\frac{89}{41}-\frac{22}{41}i$$

Multiply 5 and -1 to get -5.

$$-5-\frac{89}{41}-\frac{22}{41}i$$

Do the additions.

$$-\frac{294}{41}-\frac{22}{41}i$$

Real Part

$$-\frac{294}{41} = -7\frac{7}{41} = -7.170731707317073$$

Solution Steps

Multiply both numerator and denominator of \frac{2-3i}{4+5i} by the complex conjugate of the denominator, 4-5i.

$$Re(\frac{\left(2-3i\right)\left(4-5i\right)}{\left(4+5i\right)\left(4-5i\right)}-2+5i^{2})$$

Do the multiplications in \frac{\left(2-3i\right)\left(4-5i\right)}{\left(4+5i\right)\left(4-5i\right)}.

$$Re(\frac{-7-22i}{41}-2+5i^{2})$$

Divide -7-22i by 41 to get -\frac{7}{41}-\frac{22}{41}i.

$$Re(-\frac{7}{41}-\frac{22}{41}i-2+5i^{2})$$

Do the additions in -\frac{7}{41}-\frac{22}{41}i-2.

$$Re(5i^{2}-\frac{89}{41}-\frac{22}{41}i)$$

Calculate i to the power of 2 and get -1.

$$Re(5\left(-1\right)-\frac{89}{41}-\frac{22}{41}i)$$

Multiply 5 and -1 to get -5.

$$Re(-5-\frac{89}{41}-\frac{22}{41}i)$$

Do the additions in -5-\frac{89}{41}-\frac{22}{41}i.

$$Re(-\frac{294}{41}-\frac{22}{41}i)$$

The real part of -\frac{294}{41}-\frac{22}{41}i is -\frac{294}{41}.

$$-\frac{294}{41}$$