Solve (3 + 2/i)(i ^ 6 - i ^ 7)(i + i ^ 11 * 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

$$(3+\frac{2}{i})(i^{6}-i^{7})(i+i^{11}i)=$$

Evaluate

$-6+4i$

Solution Steps

To multiply powers of the same base, add their exponents. Add $11$ and $1$ to get $12$.

$$\left(3+\frac{2}{i}\right)\left(i^{6}-i^{7}\right)\left(i+i^{12}\right)$$

Multiply both numerator and denominator of $\frac{2}{i}$ by imaginary unit $i$.

$$\left(3+\frac{2i}{-1}\right)\left(i^{6}-i^{7}\right)\left(i+i^{12}\right)$$

Divide $2i$ by $-1$ to get $-2i$.

$$\left(3-2i\right)\left(i^{6}-i^{7}\right)\left(i+i^{12}\right)$$

Calculate $i$ to the power of $6$ and get $-1$.

$$\left(3-2i\right)\left(-1-i^{7}\right)\left(i+i^{12}\right)$$

Calculate $i$ to the power of $7$ and get $-i$.

$$\left(3-2i\right)\left(-1-\left(-i\right)\right)\left(i+i^{12}\right)$$

The opposite of $-i$ is $i$.

$$\left(3-2i\right)\left(-1+i\right)\left(i+i^{12}\right)$$

Multiply $3-2i$ and $-1+i$ to get $-1+5i$.

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

Calculate $i$ to the power of $12$ and get $1$.

$$\left(-1+5i\right)\left(i+1\right)$$

Use the distributive property to multiply $-1+5i$ by $i+1$.

$$-5-i+\left(-1+5i\right)$$

Add $-5-i$ and $-1+5i$ to get $-6+4i$.

$$-6+4i$$

Real Part

$-6$

Solution Steps

To multiply powers of the same base, add their exponents. Add $11$ and $1$ to get $12$.

$$Re(\left(3+\frac{2}{i}\right)\left(i^{6}-i^{7}\right)\left(i+i^{12}\right))$$

Multiply both numerator and denominator of $\frac{2}{i}$ by imaginary unit $i$.

$$Re(\left(3+\frac{2i}{-1}\right)\left(i^{6}-i^{7}\right)\left(i+i^{12}\right))$$

Divide $2i$ by $-1$ to get $-2i$.

$$Re(\left(3-2i\right)\left(i^{6}-i^{7}\right)\left(i+i^{12}\right))$$

Calculate $i$ to the power of $6$ and get $-1$.

$$Re(\left(3-2i\right)\left(-1-i^{7}\right)\left(i+i^{12}\right))$$

Calculate $i$ to the power of $7$ and get $-i$.

$$Re(\left(3-2i\right)\left(-1-\left(-i\right)\right)\left(i+i^{12}\right))$$

The opposite of $-i$ is $i$.

$$Re(\left(3-2i\right)\left(-1+i\right)\left(i+i^{12}\right))$$

Multiply $3-2i$ and $-1+i$ to get $-1+5i$.

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

Calculate $i$ to the power of $12$ and get $1$.

$$Re(\left(-1+5i\right)\left(i+1\right))$$

Use the distributive property to multiply $-1+5i$ by $i+1$.

$$Re(-5-i+\left(-1+5i\right))$$

Add $-5-i$ and $-1+5i$ to get $-6+4i$.

$$Re(-6+4i)$$

The real part of $-6+4i$ is $-6$.

$$-6$$