Solve ((- 2 + i) - (3 - 2i))/((- 2 + i) + (3 - 2i)) + ((- 2 + i) + (3 - 2i))/((- 2 + i)(3 - 2i)) = | 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

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

Evaluate

$-\frac{271}{65}-\frac{68}{65}i\approx -4.169230769-1.046153846i$

Solution Steps

Subtract $3-2i$ from $-2+i$ by subtracting corresponding real and imaginary parts.

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

Subtract $3$ from $-2$. Subtract $-2$ from $1$.

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

Combine the real and imaginary parts in numbers $-2+i$ and $3-2i$.

$$\frac{-5+3i}{-2+3+\left(1-2\right)i}+\frac{-2+i+\left(3-2i\right)}{\left(-2+i\right)\left(3-2i\right)}$$

Add $-2$ to $3$. Add $1$ to $-2$.

$$\frac{-5+3i}{1-i}+\frac{-2+i+\left(3-2i\right)}{\left(-2+i\right)\left(3-2i\right)}$$

Multiply both numerator and denominator of $\frac{-5+3i}{1-i}$ by the complex conjugate of the denominator, $1+i$.

$$\frac{\left(-5+3i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}+\frac{-2+i+\left(3-2i\right)}{\left(-2+i\right)\left(3-2i\right)}$$

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

$$\frac{\left(-5+3i\right)\left(1+i\right)}{1^{2}-i^{2}}+\frac{-2+i+\left(3-2i\right)}{\left(-2+i\right)\left(3-2i\right)}$$

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

$$\frac{\left(-5+3i\right)\left(1+i\right)}{2}+\frac{-2+i+\left(3-2i\right)}{\left(-2+i\right)\left(3-2i\right)}$$

Multiply complex numbers $-5+3i$ and $1+i$ like you multiply binomials.

$$\frac{-5-5i+3i\times 1+3i^{2}}{2}+\frac{-2+i+\left(3-2i\right)}{\left(-2+i\right)\left(3-2i\right)}$$

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

$$\frac{-5-5i+3i\times 1+3\left(-1\right)}{2}+\frac{-2+i+\left(3-2i\right)}{\left(-2+i\right)\left(3-2i\right)}$$

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

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

Combine the real and imaginary parts in $-5-5i+3i-3$.

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

Do the additions in $-5-3+\left(-5+3\right)i$.

$$\frac{-8-2i}{2}+\frac{-2+i+\left(3-2i\right)}{\left(-2+i\right)\left(3-2i\right)}$$

Divide $-8-2i$ by $2$ to get $-4-i$.

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

Combine the real and imaginary parts in numbers $-2+i$ and $3-2i$.

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

Add $-2$ to $3$. Add $1$ to $-2$.

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

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

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

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

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

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

$$-4-i+\frac{1-i}{-6+4i+3i+2}$$

Combine the real and imaginary parts in $-6+4i+3i+2$.

$$-4-i+\frac{1-i}{-6+2+\left(4+3\right)i}$$

Do the additions in $-6+2+\left(4+3\right)i$.

$$-4-i+\frac{1-i}{-4+7i}$$

Multiply both numerator and denominator of $\frac{1-i}{-4+7i}$ by the complex conjugate of the denominator, $-4-7i$.

$$-4-i+\frac{\left(1-i\right)\left(-4-7i\right)}{\left(-4+7i\right)\left(-4-7i\right)}$$

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

$$-4-i+\frac{\left(1-i\right)\left(-4-7i\right)}{\left(-4\right)^{2}-7^{2}i^{2}}$$

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

$$-4-i+\frac{\left(1-i\right)\left(-4-7i\right)}{65}$$

Multiply complex numbers $1-i$ and $-4-7i$ like you multiply binomials.

$$-4-i+\frac{1\left(-4\right)+1\times \left(-7i\right)-i\left(-4\right)-\left(-7i^{2}\right)}{65}$$

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

$$-4-i+\frac{1\left(-4\right)+1\times \left(-7i\right)-i\left(-4\right)-\left(-7\left(-1\right)\right)}{65}$$

Do the multiplications in $1\left(-4\right)+1\times \left(-7i\right)-i\left(-4\right)-\left(-7\left(-1\right)\right)$.

$$-4-i+\frac{-4-7i+4i-7}{65}$$

Combine the real and imaginary parts in $-4-7i+4i-7$.

$$-4-i+\frac{-4-7+\left(-7+4\right)i}{65}$$

Do the additions in $-4-7+\left(-7+4\right)i$.

$$-4-i+\frac{-11-3i}{65}$$

Divide $-11-3i$ by $65$ to get $-\frac{11}{65}-\frac{3}{65}i$.

$$-4-i+\left(-\frac{11}{65}-\frac{3}{65}i\right)$$

Combine the real and imaginary parts in numbers $-4-i$ and $-\frac{11}{65}-\frac{3}{65}i$.

$$-4-\frac{11}{65}+\left(-1-\frac{3}{65}\right)i$$

Add $-4$ to $-\frac{11}{65}$. Add $-1$ to $-\frac{3}{65}$.

$$-\frac{271}{65}-\frac{68}{65}i$$

Real Part

$-\frac{271}{65} = -4\frac{11}{65} = -4.1692307692307695$

Solution Steps

Subtract $3-2i$ from $-2+i$ by subtracting corresponding real and imaginary parts.

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

Subtract $3$ from $-2$. Subtract $-2$ from $1$.

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

Combine the real and imaginary parts in numbers $-2+i$ and $3-2i$.

$$Re(\frac{-5+3i}{-2+3+\left(1-2\right)i}+\frac{-2+i+\left(3-2i\right)}{\left(-2+i\right)\left(3-2i\right)})$$

Add $-2$ to $3$. Add $1$ to $-2$.

$$Re(\frac{-5+3i}{1-i}+\frac{-2+i+\left(3-2i\right)}{\left(-2+i\right)\left(3-2i\right)})$$

Multiply both numerator and denominator of $\frac{-5+3i}{1-i}$ by the complex conjugate of the denominator, $1+i$.

$$Re(\frac{\left(-5+3i\right)\left(1+i\right)}{\left(1-i\right)\left(1+i\right)}+\frac{-2+i+\left(3-2i\right)}{\left(-2+i\right)\left(3-2i\right)})$$

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

$$Re(\frac{\left(-5+3i\right)\left(1+i\right)}{1^{2}-i^{2}}+\frac{-2+i+\left(3-2i\right)}{\left(-2+i\right)\left(3-2i\right)})$$

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

$$Re(\frac{\left(-5+3i\right)\left(1+i\right)}{2}+\frac{-2+i+\left(3-2i\right)}{\left(-2+i\right)\left(3-2i\right)})$$

Multiply complex numbers $-5+3i$ and $1+i$ like you multiply binomials.

$$Re(\frac{-5-5i+3i\times 1+3i^{2}}{2}+\frac{-2+i+\left(3-2i\right)}{\left(-2+i\right)\left(3-2i\right)})$$

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

$$Re(\frac{-5-5i+3i\times 1+3\left(-1\right)}{2}+\frac{-2+i+\left(3-2i\right)}{\left(-2+i\right)\left(3-2i\right)})$$

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

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

Combine the real and imaginary parts in $-5-5i+3i-3$.

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

Do the additions in $-5-3+\left(-5+3\right)i$.

$$Re(\frac{-8-2i}{2}+\frac{-2+i+\left(3-2i\right)}{\left(-2+i\right)\left(3-2i\right)})$$

Divide $-8-2i$ by $2$ to get $-4-i$.

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

Combine the real and imaginary parts in numbers $-2+i$ and $3-2i$.

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

Add $-2$ to $3$. Add $1$ to $-2$.

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

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

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

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

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

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

$$Re(-4-i+\frac{1-i}{-6+4i+3i+2})$$

Combine the real and imaginary parts in $-6+4i+3i+2$.

$$Re(-4-i+\frac{1-i}{-6+2+\left(4+3\right)i})$$

Do the additions in $-6+2+\left(4+3\right)i$.

$$Re(-4-i+\frac{1-i}{-4+7i})$$

Multiply both numerator and denominator of $\frac{1-i}{-4+7i}$ by the complex conjugate of the denominator, $-4-7i$.

$$Re(-4-i+\frac{\left(1-i\right)\left(-4-7i\right)}{\left(-4+7i\right)\left(-4-7i\right)})$$

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

$$Re(-4-i+\frac{\left(1-i\right)\left(-4-7i\right)}{\left(-4\right)^{2}-7^{2}i^{2}})$$

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

$$Re(-4-i+\frac{\left(1-i\right)\left(-4-7i\right)}{65})$$

Multiply complex numbers $1-i$ and $-4-7i$ like you multiply binomials.

$$Re(-4-i+\frac{1\left(-4\right)+1\times \left(-7i\right)-i\left(-4\right)-\left(-7i^{2}\right)}{65})$$

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

$$Re(-4-i+\frac{1\left(-4\right)+1\times \left(-7i\right)-i\left(-4\right)-\left(-7\left(-1\right)\right)}{65})$$

Do the multiplications in $1\left(-4\right)+1\times \left(-7i\right)-i\left(-4\right)-\left(-7\left(-1\right)\right)$.

$$Re(-4-i+\frac{-4-7i+4i-7}{65})$$

Combine the real and imaginary parts in $-4-7i+4i-7$.

$$Re(-4-i+\frac{-4-7+\left(-7+4\right)i}{65})$$

Do the additions in $-4-7+\left(-7+4\right)i$.

$$Re(-4-i+\frac{-11-3i}{65})$$

Divide $-11-3i$ by $65$ to get $-\frac{11}{65}-\frac{3}{65}i$.

$$Re(-4-i+\left(-\frac{11}{65}-\frac{3}{65}i\right))$$

Combine the real and imaginary parts in numbers $-4-i$ and $-\frac{11}{65}-\frac{3}{65}i$.

$$Re(-4-\frac{11}{65}+\left(-1-\frac{3}{65}\right)i)$$

Add $-4$ to $-\frac{11}{65}$. Add $-1$ to $-\frac{3}{65}$.

$$Re(-\frac{271}{65}-\frac{68}{65}i)$$

The real part of $-\frac{271}{65}-\frac{68}{65}i$ is $-\frac{271}{65}$.

$$-\frac{271}{65}$$