Answer
$-4$
Work Step by Step
Note: the standard form of a complex number is $a+bi$ where $i = \sqrt -1$ and $a$ and $b$ are real numbers.
Note: $i^{2} = \sqrt -1\sqrt -1 = -1$
$(1+i)^{4}$
The complex number can be rewritten as:
$(1+i)^{4}$ = $(1+i)^{2}(1+i)^{2}$
Solve for $(1+i)^{2}$:
$(1+i)^{2} = (1+i)(1+i)$
Expand out the complex number:
$(1+i)^{2} = 1+i+i+i^{2}$
As seen in the note above, $i^{2} = -1$, so substitute in $-1$.
$(1+i)^{2} = 1+i+i+(-1)$
Simplify:
$(1+i)^{2} = 1+i+i-1$
Combine like terms, to express the complex number in standard form:
$(1+i)^{2} = 2i$
Since $(1+i)^{2} = 2i$, substitute into the expression for $(1+i)^{4}$ from above:
$(1+i)^{4}$ = $(1+i)^{2}(1+i)^{2}$
$(1+i)^{4}=(2i)(2i)$
$(1+i)^{4} = 4i^{2}$
As seen in the note above, $i^{2} = -1$, so substitute in $-1$.
$(1+i)^{4} = 4(-1)$
$(1+i)^{4} = -4$
