Solve z=(a-3)i^3+(a+4)i^6

$$z = ( a - 3 ) i ^ { 3 } + ( a + 4 ) i ^ { 6 }$$

Solve for a

$$a=\left(-\frac{1}{2}+\frac{1}{2}i\right)z+\left(-\frac{1}{2}+\frac{7}{2}i\right)$$

Steps for Solving Linear Equation

Calculate i to the power of 3 and get -i.

$$z=\left(a-3\right)\left(-i\right)+\left(a+4\right)i^{6}$$

Use the distributive property to multiply a-3 by -i.

$$z=-ia+3i+\left(a+4\right)i^{6}$$

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

$$z=-ia+3i+\left(a+4\right)\left(-1\right)$$

Use the distributive property to multiply a+4 by -1.

$$z=-ia+3i-a-4$$

Combine -ia and -a to get \left(-1-i\right)a.

$$z=\left(-1-i\right)a+3i-4$$

Swap sides so that all variable terms are on the left hand side.

$$\left(-1-i\right)a+3i-4=z$$

Subtract 3i from both sides.

$$\left(-1-i\right)a-4=z-3i$$

Add 4 to both sides.

$$\left(-1-i\right)a=z-3i+4$$

The equation is in standard form.

$$\left(-1-i\right)a=z+\left(4-3i\right)$$

Divide both sides by -1-i.

$$\frac{\left(-1-i\right)a}{-1-i}=\frac{z+\left(4-3i\right)}{-1-i}$$

Dividing by -1-i undoes the multiplication by -1-i.

$$a=\frac{z+\left(4-3i\right)}{-1-i}$$

Divide z+\left(4-3i\right) by -1-i.

$$a=\left(-\frac{1}{2}+\frac{1}{2}i\right)z+\left(-\frac{1}{2}+\frac{7}{2}i\right)$$

Solve for z

$$z=\left(-1-i\right)a+\left(-4+3i\right)$$