Answer
$-95 +63i$
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: $i^{2} = \sqrt -1\sqrt -1 = -1$
$(3+4i)^{2}(4+3i) - (2+3i)^{2}$
Solve for $(3+4i)^{2}$:
$(3+4i)^{2} = (3+4i)(3+4i)$
Expand out the complex number:
$(3+4i)^{2} = 9+12i+12i+16i^{2}$
As seen in the note above, $i^{2} = -1$, so substitute in $-1$.
$(3+4i)^{2} = 9+12i+12i+16(-1)$
Simplify:
$(3+4i)^{2} = 9+12i+12i-16$
Combine like terms, to express the complex number in standard form:
$(3+4i)^{2} = -7+24i$
Solve for $(2+3i)^{2}$:
$(2+3i)^{2} = (2+3i)(2+3i)$
Expand out the complex number:
$(2+3i)^{2} = 4 + 6i + 6i + 9i^{2}$
As seen in the note above, $i^{2} = -1$, so substitute in $-1$.
$(2+3i)^{2} = 4 + 6i + 6i + 9(-1)$
Simplify:
$(2+3i)^{2} = 4 + 6i + 6i -9$
Combine like terms, to express the complex number in standard form:
$(2+3i)^{2} = -5+12i$
Since $(3+4i)^{2} = -7+24i$ and $(2+3i)^{2} = -5+12i$, substitute into $(3+4i)^{2}(4+3i) - (2+3i)^{2}$:
$(3+4i)^{2}(4+3i) - (2+3i)^{2} = ( -7+24i)(4+3i)- (-5+12i)$
Expand out the complex number:
$(3+4i)^{2}(4+3i) - (2+3i)^{2} = (-28-21i+96i+72i^{2})- (-5+12i)$
As seen in the note above, $i^{2} = -1$, so substitute in $-1$.
$(3+4i)^{2}(4+3i) - (2+3i)^{2} = (-28-21i+96i+72(-1))- (-5+12i)$
Simplify:
$(3+4i)^{2}(4+3i) - (2+3i)^{2} = (-28-21i+96i-72)- (-5+12i)$
$(3+4i)^{2}(4+3i) - (2+3i)^{2} = -28-21i+96i-72 + 5 -12i$
Combine like terms, to express the complex number in standard form:
$(3+4i)^{2}(4+3i) - (2+3i)^{2} = -95 +63i$
