Solve (3-26)/(4+3i)*(4-36)/(4-3i)

$$\frac { ( 3 - 26 ) } { ( 4 + 3 i ) } \times \frac { ( 4 - 36 ) } { ( 4 - 3 i ) }$$

Evaluate

$$\frac{736}{25}=29.44$$

Solution Steps

Subtract 26 from 3 to get -23.

$$\frac{-23}{4+3i}\times \frac{4-36}{4-3i}$$

Multiply both numerator and denominator of \frac{-23}{4+3i} by the complex conjugate of the denominator, 4-3i.

$$\frac{-23\left(4-3i\right)}{\left(4+3i\right)\left(4-3i\right)}\times \frac{4-36}{4-3i}$$

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

$$\frac{-23\left(4-3i\right)}{4^{2}-3^{2}i^{2}}\times \frac{4-36}{4-3i}$$

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

$$\frac{-23\left(4-3i\right)}{25}\times \frac{4-36}{4-3i}$$

Multiply -23 times 4-3i.

$$\frac{-23\times 4-23\times \left(-3i\right)}{25}\times \frac{4-36}{4-3i}$$

Do the multiplications in -23\times 4-23\times \left(-3i\right).

$$\frac{-92+69i}{25}\times \frac{4-36}{4-3i}$$

Divide -92+69i by 25 to get -\frac{92}{25}+\frac{69}{25}i.

$$\left(-\frac{92}{25}+\frac{69}{25}i\right)\times \frac{4-36}{4-3i}$$

Subtract 36 from 4 to get -32.

$$\left(-\frac{92}{25}+\frac{69}{25}i\right)\times \frac{-32}{4-3i}$$

Multiply both numerator and denominator of \frac{-32}{4-3i} by the complex conjugate of the denominator, 4+3i.

$$\left(-\frac{92}{25}+\frac{69}{25}i\right)\times \frac{-32\left(4+3i\right)}{\left(4-3i\right)\left(4+3i\right)}$$

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

$$\left(-\frac{92}{25}+\frac{69}{25}i\right)\times \frac{-32\left(4+3i\right)}{4^{2}-3^{2}i^{2}}$$

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

$$\left(-\frac{92}{25}+\frac{69}{25}i\right)\times \frac{-32\left(4+3i\right)}{25}$$

Multiply -32 times 4+3i.

$$\left(-\frac{92}{25}+\frac{69}{25}i\right)\times \frac{-32\times 4-32\times \left(3i\right)}{25}$$

Do the multiplications in -32\times 4-32\times \left(3i\right).

$$\left(-\frac{92}{25}+\frac{69}{25}i\right)\times \frac{-128-96i}{25}$$

Divide -128-96i by 25 to get -\frac{128}{25}-\frac{96}{25}i.

$$\left(-\frac{92}{25}+\frac{69}{25}i\right)\left(-\frac{128}{25}-\frac{96}{25}i\right)$$

Multiply complex numbers -\frac{92}{25}+\frac{69}{25}i and -\frac{128}{25}-\frac{96}{25}i like you multiply binomials.

$$-\frac{92}{25}\left(-\frac{128}{25}\right)-\frac{92}{25}\times \left(-\frac{96}{25}i\right)+\frac{69}{25}i\left(-\frac{128}{25}\right)+\frac{69}{25}\left(-\frac{96}{25}\right)i^{2}$$

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

$$-\frac{92}{25}\left(-\frac{128}{25}\right)-\frac{92}{25}\times \left(-\frac{96}{25}i\right)+\frac{69}{25}i\left(-\frac{128}{25}\right)+\frac{69}{25}\left(-\frac{96}{25}\right)\left(-1\right)$$

Do the multiplications.

$$\frac{11776}{625}+\frac{8832}{625}i-\frac{8832}{625}i+\frac{6624}{625}$$

Combine the real and imaginary parts.

$$\frac{11776}{625}+\frac{6624}{625}+\left(\frac{8832}{625}-\frac{8832}{625}\right)i$$

Do the additions.

$$\frac{736}{25}$$

Real Part

$$\frac{736}{25} = 29\frac{11}{25} = 29.44$$

Solution Steps

Subtract 26 from 3 to get -23.

$$Re(\frac{-23}{4+3i}\times \frac{4-36}{4-3i})$$

Multiply both numerator and denominator of \frac{-23}{4+3i} by the complex conjugate of the denominator, 4-3i.

$$Re(\frac{-23\left(4-3i\right)}{\left(4+3i\right)\left(4-3i\right)}\times \frac{4-36}{4-3i})$$

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

$$Re(\frac{-23\left(4-3i\right)}{4^{2}-3^{2}i^{2}}\times \frac{4-36}{4-3i})$$

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

$$Re(\frac{-23\left(4-3i\right)}{25}\times \frac{4-36}{4-3i})$$

Multiply -23 times 4-3i.

$$Re(\frac{-23\times 4-23\times \left(-3i\right)}{25}\times \frac{4-36}{4-3i})$$

Do the multiplications in -23\times 4-23\times \left(-3i\right).

$$Re(\frac{-92+69i}{25}\times \frac{4-36}{4-3i})$$

Divide -92+69i by 25 to get -\frac{92}{25}+\frac{69}{25}i.

$$Re(\left(-\frac{92}{25}+\frac{69}{25}i\right)\times \frac{4-36}{4-3i})$$

Subtract 36 from 4 to get -32.

$$Re(\left(-\frac{92}{25}+\frac{69}{25}i\right)\times \frac{-32}{4-3i})$$

Multiply both numerator and denominator of \frac{-32}{4-3i} by the complex conjugate of the denominator, 4+3i.

$$Re(\left(-\frac{92}{25}+\frac{69}{25}i\right)\times \frac{-32\left(4+3i\right)}{\left(4-3i\right)\left(4+3i\right)})$$

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

$$Re(\left(-\frac{92}{25}+\frac{69}{25}i\right)\times \frac{-32\left(4+3i\right)}{4^{2}-3^{2}i^{2}})$$

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

$$Re(\left(-\frac{92}{25}+\frac{69}{25}i\right)\times \frac{-32\left(4+3i\right)}{25})$$

Multiply -32 times 4+3i.

$$Re(\left(-\frac{92}{25}+\frac{69}{25}i\right)\times \frac{-32\times 4-32\times \left(3i\right)}{25})$$

Do the multiplications in -32\times 4-32\times \left(3i\right).

$$Re(\left(-\frac{92}{25}+\frac{69}{25}i\right)\times \frac{-128-96i}{25})$$

Divide -128-96i by 25 to get -\frac{128}{25}-\frac{96}{25}i.

$$Re(\left(-\frac{92}{25}+\frac{69}{25}i\right)\left(-\frac{128}{25}-\frac{96}{25}i\right))$$

Multiply complex numbers -\frac{92}{25}+\frac{69}{25}i and -\frac{128}{25}-\frac{96}{25}i like you multiply binomials.

$$Re(-\frac{92}{25}\left(-\frac{128}{25}\right)-\frac{92}{25}\times \left(-\frac{96}{25}i\right)+\frac{69}{25}i\left(-\frac{128}{25}\right)+\frac{69}{25}\left(-\frac{96}{25}\right)i^{2})$$

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

$$Re(-\frac{92}{25}\left(-\frac{128}{25}\right)-\frac{92}{25}\times \left(-\frac{96}{25}i\right)+\frac{69}{25}i\left(-\frac{128}{25}\right)+\frac{69}{25}\left(-\frac{96}{25}\right)\left(-1\right))$$

Do the multiplications in -\frac{92}{25}\left(-\frac{128}{25}\right)-\frac{92}{25}\times \left(-\frac{96}{25}i\right)+\frac{69}{25}i\left(-\frac{128}{25}\right)+\frac{69}{25}\left(-\frac{96}{25}\right)\left(-1\right).

$$Re(\frac{11776}{625}+\frac{8832}{625}i-\frac{8832}{625}i+\frac{6624}{625})$$

Combine the real and imaginary parts in \frac{11776}{625}+\frac{8832}{625}i-\frac{8832}{625}i+\frac{6624}{625}.

$$Re(\frac{11776}{625}+\frac{6624}{625}+\left(\frac{8832}{625}-\frac{8832}{625}\right)i)$$

Do the additions in \frac{11776}{625}+\frac{6624}{625}+\left(\frac{8832}{625}-\frac{8832}{625}\right)i.

$$Re(\frac{736}{25})$$

The real part of \frac{736}{25} is \frac{736}{25}.

$$\frac{736}{25}$$