Solve 4i/6.4+4i(3sqrt{2}i)

$$\frac { 4 i } { 6.4 + 4 i } ( 3 \sqrt { 2 } i )$$

Evaluate

$$\left(-\frac{120}{89}+\frac{75}{89}i\right)\sqrt{2}\approx -1.906804803+1.191753002i$$

Solution Steps

Multiply both numerator and denominator of \frac{4i}{6.4+4i} by the complex conjugate of the denominator, 6.4-4i.

$$\frac{4i\left(6.4-4i\right)}{\left(6.4+4i\right)\left(6.4-4i\right)}\times \left(3i\right)\sqrt{2}$$

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

$$\frac{4i\left(6.4-4i\right)}{6.4^{2}-4^{2}i^{2}}\times \left(3i\right)\sqrt{2}$$

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

$$\frac{4i\left(6.4-4i\right)}{56.96}\times \left(3i\right)\sqrt{2}$$

Multiply 4i times 6.4-4i.

$$\frac{4i\times 6.4+4\left(-4\right)i^{2}}{56.96}\times \left(3i\right)\sqrt{2}$$

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

$$\frac{4i\times 6.4+4\left(-4\right)\left(-1\right)}{56.96}\times \left(3i\right)\sqrt{2}$$

Do the multiplications in 4i\times 6.4+4\left(-4\right)\left(-1\right). Reorder the terms.

$$\frac{16+25.6i}{56.96}\times \left(3i\right)\sqrt{2}$$

Divide 16+25.6i by 56.96 to get \frac{25}{89}+\frac{40}{89}i.

$$\left(\frac{25}{89}+\frac{40}{89}i\right)\times \left(3i\right)\sqrt{2}$$

Multiply \frac{25}{89}+\frac{40}{89}i times 3i.

$$\left(\frac{25}{89}\times \left(3i\right)+\frac{40}{89}\times 3i^{2}\right)\sqrt{2}$$

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

$$\left(\frac{25}{89}\times \left(3i\right)+\frac{40}{89}\times 3\left(-1\right)\right)\sqrt{2}$$

Do the multiplications. Reorder the terms.

$$\left(-\frac{120}{89}+\frac{75}{89}i\right)\sqrt{2}$$

Real Part

$$-\frac{120 \sqrt{2}}{89} = -1.9068048031996787$$