Solve 23*2-3i/4+3i

$$23 \cdot \frac { 2 - 3 i } { 4 + 3 i }$$

Evaluate

$$-\frac{23}{25}-\frac{414}{25}i=-0.92-16.56i$$

Solution Steps

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

$$23\times \frac{\left(2-3i\right)\left(4-3i\right)}{\left(4+3i\right)\left(4-3i\right)}$$

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

$$23\times \frac{\left(2-3i\right)\left(4-3i\right)}{4^{2}-3^{2}i^{2}}$$

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

$$23\times \frac{\left(2-3i\right)\left(4-3i\right)}{25}$$

Multiply complex numbers 2-3i and 4-3i like you multiply binomials.

$$23\times \frac{2\times 4+2\times \left(-3i\right)-3i\times 4-3\left(-3\right)i^{2}}{25}$$

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

$$23\times \frac{2\times 4+2\times \left(-3i\right)-3i\times 4-3\left(-3\right)\left(-1\right)}{25}$$

Do the multiplications in 2\times 4+2\times \left(-3i\right)-3i\times 4-3\left(-3\right)\left(-1\right).

$$23\times \frac{8-6i-12i-9}{25}$$

Combine the real and imaginary parts in 8-6i-12i-9.

$$23\times \frac{8-9+\left(-6-12\right)i}{25}$$

Do the additions in 8-9+\left(-6-12\right)i.

$$23\times \frac{-1-18i}{25}$$

Divide -1-18i by 25 to get -\frac{1}{25}-\frac{18}{25}i.

$$23\left(-\frac{1}{25}-\frac{18}{25}i\right)$$

Multiply 23 times -\frac{1}{25}-\frac{18}{25}i.

$$23\left(-\frac{1}{25}\right)+23\times \left(-\frac{18}{25}i\right)$$

Do the multiplications.

$$-\frac{23}{25}-\frac{414}{25}i$$

Real Part

$$-\frac{23}{25} = -0.92$$

Solution Steps

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

$$Re(23\times \frac{\left(2-3i\right)\left(4-3i\right)}{\left(4+3i\right)\left(4-3i\right)})$$

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

$$Re(23\times \frac{\left(2-3i\right)\left(4-3i\right)}{4^{2}-3^{2}i^{2}})$$

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

$$Re(23\times \frac{\left(2-3i\right)\left(4-3i\right)}{25})$$

Multiply complex numbers 2-3i and 4-3i like you multiply binomials.

$$Re(23\times \frac{2\times 4+2\times \left(-3i\right)-3i\times 4-3\left(-3\right)i^{2}}{25})$$

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

$$Re(23\times \frac{2\times 4+2\times \left(-3i\right)-3i\times 4-3\left(-3\right)\left(-1\right)}{25})$$

Do the multiplications in 2\times 4+2\times \left(-3i\right)-3i\times 4-3\left(-3\right)\left(-1\right).

$$Re(23\times \frac{8-6i-12i-9}{25})$$

Combine the real and imaginary parts in 8-6i-12i-9.

$$Re(23\times \frac{8-9+\left(-6-12\right)i}{25})$$

Do the additions in 8-9+\left(-6-12\right)i.

$$Re(23\times \frac{-1-18i}{25})$$

Divide -1-18i by 25 to get -\frac{1}{25}-\frac{18}{25}i.

$$Re(23\left(-\frac{1}{25}-\frac{18}{25}i\right))$$

Multiply 23 times -\frac{1}{25}-\frac{18}{25}i.

$$Re(23\left(-\frac{1}{25}\right)+23\times \left(-\frac{18}{25}i\right))$$

Do the multiplications in 23\left(-\frac{1}{25}\right)+23\times \left(-\frac{18}{25}i\right).

$$Re(-\frac{23}{25}-\frac{414}{25}i)$$

The real part of -\frac{23}{25}-\frac{414}{25}i is -\frac{23}{25}.

$$-\frac{23}{25}$$