Solve i^-1*3x(-3)^-2+(-1/9)^0/2^-2-1

$$\frac { i ^ { - 1 } \cdot 3 x ( - 3 ) ^ { - 2 } + ( - \frac { 1 } { 9 } ) ^ { 0 } } { 2 ^ { - 2 } - 1 }$$

Evaluate

$$\frac{4ix}{9}-\frac{4}{3}$$

Solution Steps

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

$$\frac{-i\times 3x\left(-3\right)^{-2}+\left(-\frac{1}{9}\right)^{0}}{2^{-2}-1}$$

Multiply -i and 3 to get -3i.

$$\frac{-3ix\left(-3\right)^{-2}+\left(-\frac{1}{9}\right)^{0}}{2^{-2}-1}$$

Calculate -3 to the power of -2 and get \frac{1}{9}.

$$\frac{-3ix\times \frac{1}{9}+\left(-\frac{1}{9}\right)^{0}}{2^{-2}-1}$$

Multiply -3i and \frac{1}{9} to get -\frac{1}{3}i.

$$\frac{-\frac{1}{3}ix+\left(-\frac{1}{9}\right)^{0}}{2^{-2}-1}$$

Calculate -\frac{1}{9} to the power of 0 and get 1.

$$\frac{-\frac{1}{3}ix+1}{2^{-2}-1}$$

Calculate 2 to the power of -2 and get \frac{1}{4}.

$$\frac{-\frac{1}{3}ix+1}{\frac{1}{4}-1}$$

Subtract 1 from \frac{1}{4} to get -\frac{3}{4}.

$$\frac{-\frac{1}{3}ix+1}{-\frac{3}{4}}$$

Divide -\frac{1}{3}ix+1 by -\frac{3}{4} by multiplying -\frac{1}{3}ix+1 by the reciprocal of -\frac{3}{4}.

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

Use the distributive property to multiply -\frac{1}{3}ix+1 by 4.

$$\frac{-\frac{4}{3}ix+4}{-3}$$

Multiply both numerator and denominator by -1.

$$\frac{\frac{4}{3}ix-4}{3}$$