Value of (1 + i)⁸, where $i=\sqrt{-1}$, is equal to

The correct answer is 16.

To find the value of ${(1+i)}^8$, we can first calculate the square of the expression, ${(1+i)}^2$, and then raise the result to the 4th power.

Using the algebraic identity ${(a+b)}^2=a^2+2ab+b^2$, we expand ${(1+i)}^2$:
${(1+i)}^2=1^2+2\left(1\right)\left(i\right)+i^2$

Since $i=\sqrt{-1}$, we know that $i^2=-1$. Substituting this value into the expression, we get:
${(1+i)}^2=1+2i-1$
${(1+i)}^2=2i$

Now, we can express ${(1+i)}^8$ using the law of exponents as:
${(1+i)}^8={\left[{(1+i)}^2\right]}^4$

Substitute ${(1+i)}^2=2i$ into the equation:
${(1+i)}^8={\left(2i\right)}^4$
${(1+i)}^8=2^4\times i^4$

We compute each part:
1. $2^4=16$
2. $i^4={\left(i^2\right)}^2={(-1)}^2=1$

Therefore, we get:
${(1+i)}^8=16\times 1=16$