Solve 2-3i/frac{17-6i{25}}

$$\frac { 2 - 3 i } { \frac { 17 - 6 i } { 25 } }$$

Evaluate

$$4-3i$$

Solution Steps

Divide 2-3i by \frac{17-6i}{25} by multiplying 2-3i by the reciprocal of \frac{17-6i}{25}.

$$\frac{\left(2-3i\right)\times 25}{17-6i}$$

Multiply 2-3i times 25.

$$\frac{2\times 25-3i\times 25}{17-6i}$$

Do the multiplications in 2\times 25-3i\times 25.

$$\frac{50-75i}{17-6i}$$

Multiply both numerator and denominator by the complex conjugate of the denominator, 17+6i.

$$\frac{\left(50-75i\right)\left(17+6i\right)}{\left(17-6i\right)\left(17+6i\right)}$$

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

$$\frac{\left(50-75i\right)\left(17+6i\right)}{17^{2}-6^{2}i^{2}}$$

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

$$\frac{\left(50-75i\right)\left(17+6i\right)}{325}$$

Multiply complex numbers 50-75i and 17+6i like you multiply binomials.

$$\frac{50\times 17+50\times \left(6i\right)-75i\times 17-75\times 6i^{2}}{325}$$

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

$$\frac{50\times 17+50\times \left(6i\right)-75i\times 17-75\times 6\left(-1\right)}{325}$$

Do the multiplications in 50\times 17+50\times \left(6i\right)-75i\times 17-75\times 6\left(-1\right).

$$\frac{850+300i-1275i+450}{325}$$

Combine the real and imaginary parts in 850+300i-1275i+450.

$$\frac{850+450+\left(300-1275\right)i}{325}$$

Do the additions in 850+450+\left(300-1275\right)i.

$$\frac{1300-975i}{325}$$

Divide 1300-975i by 325 to get 4-3i.

$$4-3i$$

Real Part

$$4$$

Solution Steps

Divide 2-3i by \frac{17-6i}{25} by multiplying 2-3i by the reciprocal of \frac{17-6i}{25}.

$$Re(\frac{\left(2-3i\right)\times 25}{17-6i})$$

Multiply 2-3i times 25.

$$Re(\frac{2\times 25-3i\times 25}{17-6i})$$

Do the multiplications in 2\times 25-3i\times 25.

$$Re(\frac{50-75i}{17-6i})$$

Multiply both numerator and denominator of \frac{50-75i}{17-6i} by the complex conjugate of the denominator, 17+6i.

$$Re(\frac{\left(50-75i\right)\left(17+6i\right)}{\left(17-6i\right)\left(17+6i\right)})$$

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

$$Re(\frac{\left(50-75i\right)\left(17+6i\right)}{17^{2}-6^{2}i^{2}})$$

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

$$Re(\frac{\left(50-75i\right)\left(17+6i\right)}{325})$$

Multiply complex numbers 50-75i and 17+6i like you multiply binomials.

$$Re(\frac{50\times 17+50\times \left(6i\right)-75i\times 17-75\times 6i^{2}}{325})$$

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

$$Re(\frac{50\times 17+50\times \left(6i\right)-75i\times 17-75\times 6\left(-1\right)}{325})$$

Do the multiplications in 50\times 17+50\times \left(6i\right)-75i\times 17-75\times 6\left(-1\right).

$$Re(\frac{850+300i-1275i+450}{325})$$

Combine the real and imaginary parts in 850+300i-1275i+450.

$$Re(\frac{850+450+\left(300-1275\right)i}{325})$$

Do the additions in 850+450+\left(300-1275\right)i.

$$Re(\frac{1300-975i}{325})$$

Divide 1300-975i by 325 to get 4-3i.

$$Re(4-3i)$$

The real part of 4-3i is 4.

$$4$$