Solve find the Cojuctive = 5 ((3 - 2i)(2 + 3i ^ 3))/((1 + 2i)(2 - i)) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\frac { ( 3 - 2 i ) ( 2 + 3 i ) } { ( 1 + 2 i ) ( 2 - i ) }$$

Evaluate

$\frac{63}{25}-\frac{16}{25}i=2.52-0.64i$

Solution Steps

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

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

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

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

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

$$\frac{6+9i-4i+6}{\left(1+2i\right)\left(2-i\right)}$$

Combine the real and imaginary parts in $6+9i-4i+6$.

$$\frac{6+6+\left(9-4\right)i}{\left(1+2i\right)\left(2-i\right)}$$

Do the additions in $6+6+\left(9-4\right)i$.

$$\frac{12+5i}{\left(1+2i\right)\left(2-i\right)}$$

Multiply complex numbers $1+2i$ and $2-i$ like you multiply binomials.

$$\frac{12+5i}{1\times 2+1\left(-i\right)+2i\times 2+2\left(-1\right)i^{2}}$$

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

$$\frac{12+5i}{1\times 2+1\left(-i\right)+2i\times 2+2\left(-1\right)\left(-1\right)}$$

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

$$\frac{12+5i}{2-i+4i+2}$$

Combine the real and imaginary parts in $2-i+4i+2$.

$$\frac{12+5i}{2+2+\left(-1+4\right)i}$$

Do the additions in $2+2+\left(-1+4\right)i$.

$$\frac{12+5i}{4+3i}$$

Multiply both numerator and denominator by the complex conjugate of the denominator, $4-3i$.

$$\frac{\left(12+5i\right)\left(4-3i\right)}{\left(4+3i\right)\left(4-3i\right)}$$

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

$$\frac{\left(12+5i\right)\left(4-3i\right)}{4^{2}-3^{2}i^{2}}$$

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

$$\frac{\left(12+5i\right)\left(4-3i\right)}{25}$$

Multiply complex numbers $12+5i$ and $4-3i$ like you multiply binomials.

$$\frac{12\times 4+12\times \left(-3i\right)+5i\times 4+5\left(-3\right)i^{2}}{25}$$

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

$$\frac{12\times 4+12\times \left(-3i\right)+5i\times 4+5\left(-3\right)\left(-1\right)}{25}$$

Do the multiplications in $12\times 4+12\times \left(-3i\right)+5i\times 4+5\left(-3\right)\left(-1\right)$.

$$\frac{48-36i+20i+15}{25}$$

Combine the real and imaginary parts in $48-36i+20i+15$.

$$\frac{48+15+\left(-36+20\right)i}{25}$$

Do the additions in $48+15+\left(-36+20\right)i$.

$$\frac{63-16i}{25}$$

Divide $63-16i$ by $25$ to get $\frac{63}{25}-\frac{16}{25}i$.

$$\frac{63}{25}-\frac{16}{25}i$$

Real Part

$\frac{63}{25} = 2\frac{13}{25} = 2.52$

Solution Steps

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

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

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

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

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

$$Re(\frac{6+9i-4i+6}{\left(1+2i\right)\left(2-i\right)})$$

Combine the real and imaginary parts in $6+9i-4i+6$.

$$Re(\frac{6+6+\left(9-4\right)i}{\left(1+2i\right)\left(2-i\right)})$$

Do the additions in $6+6+\left(9-4\right)i$.

$$Re(\frac{12+5i}{\left(1+2i\right)\left(2-i\right)})$$

Multiply complex numbers $1+2i$ and $2-i$ like you multiply binomials.

$$Re(\frac{12+5i}{1\times 2+1\left(-i\right)+2i\times 2+2\left(-1\right)i^{2}})$$

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

$$Re(\frac{12+5i}{1\times 2+1\left(-i\right)+2i\times 2+2\left(-1\right)\left(-1\right)})$$

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

$$Re(\frac{12+5i}{2-i+4i+2})$$

Combine the real and imaginary parts in $2-i+4i+2$.

$$Re(\frac{12+5i}{2+2+\left(-1+4\right)i})$$

Do the additions in $2+2+\left(-1+4\right)i$.

$$Re(\frac{12+5i}{4+3i})$$

Multiply both numerator and denominator of $\frac{12+5i}{4+3i}$ by the complex conjugate of the denominator, $4-3i$.

$$Re(\frac{\left(12+5i\right)\left(4-3i\right)}{\left(4+3i\right)\left(4-3i\right)})$$

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

$$Re(\frac{\left(12+5i\right)\left(4-3i\right)}{4^{2}-3^{2}i^{2}})$$

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

$$Re(\frac{\left(12+5i\right)\left(4-3i\right)}{25})$$

Multiply complex numbers $12+5i$ and $4-3i$ like you multiply binomials.

$$Re(\frac{12\times 4+12\times \left(-3i\right)+5i\times 4+5\left(-3\right)i^{2}}{25})$$

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

$$Re(\frac{12\times 4+12\times \left(-3i\right)+5i\times 4+5\left(-3\right)\left(-1\right)}{25})$$

Do the multiplications in $12\times 4+12\times \left(-3i\right)+5i\times 4+5\left(-3\right)\left(-1\right)$.

$$Re(\frac{48-36i+20i+15}{25})$$

Combine the real and imaginary parts in $48-36i+20i+15$.

$$Re(\frac{48+15+\left(-36+20\right)i}{25})$$

Do the additions in $48+15+\left(-36+20\right)i$.

$$Re(\frac{63-16i}{25})$$

Divide $63-16i$ by $25$ to get $\frac{63}{25}-\frac{16}{25}i$.

$$Re(\frac{63}{25}-\frac{16}{25}i)$$

The real part of $\frac{63}{25}-\frac{16}{25}i$ is $\frac{63}{25}$.

$$\frac{63}{25}$$