Solve operatorname{ne}(z-i-1)^3=8i

$$\operatorname { ne } ( z - i - 1 ) ^ { 3 } = 8 i$$

Solve for n

$$n=\frac{8i}{e\left(z+\left(-1-i\right)\right)^{3}}$$

$$z\neq 1+i$$

Steps for Solving Linear Equation

The equation is in standard form.

$$e\left(z+\left(-1-i\right)\right)^{3}n=8i$$

Divide both sides by e\left(z+\left(-1-i\right)\right)^{3}.

$$\frac{e\left(z+\left(-1-i\right)\right)^{3}n}{e\left(z+\left(-1-i\right)\right)^{3}}=\frac{8i}{e\left(z+\left(-1-i\right)\right)^{3}}$$

Dividing by e\left(z+\left(-1-i\right)\right)^{3} undoes the multiplication by e\left(z+\left(-1-i\right)\right)^{3}.

$$n=\frac{8i}{e\left(z+\left(-1-i\right)\right)^{3}}$$

Solve for z

$$z=2e^{\frac{\pi i}{6}-\frac{1}{3}}n^{-\frac{1}{3}}+\left(1+i\right)$$

$$z=2ie^{\frac{-1+\pi i}{3}}n^{-\frac{1}{3}}+\left(1+i\right)$$

$$z=1+i-2e^{\frac{\pi i}{2}-\frac{1}{3}}n^{-\frac{1}{3}}\text{, }n\neq 0$$