Question 61330
{{{ (c-2i)^5 }}}
let's expand this out
{{{ (c-2i)(c-2i)(c-2i)(c-2i)(c-2i) }}}
Looking at just the first 2 quantities ... let's FOIL them
F: {{{ c * c = c^2 }}}
O: {{{ c * -2i = -2ci }}}
I: {{{ -2i * c = -2ci }}}
L: {{{ -2i * -2i = 4i^2 =4(-1) = -4 }}}
remember i^2 = -1
Combine like terms
{{{ c^2 -4ci -4 }}}
Now the 3rd and 4th terms when FOILed, give the same response.
You now have the following
{{{ (c^2 -4ci -4)(c^2 -4ci -4)(c-2i) }}}
now multiply out the two trinomials
{{{ (c^2*c^2)+(c^2*-4ci)+(c^2*-4)+(-4ci*c^2)+(-4ci*-4ci)+(-4ci*-4)+(-4*c^2)+(-4*-4ci)+(-4*-4) }}}
simplify
{{{ c^4 -4c^3i -4c^2 -4c^3i -16c^2i^2 +16ci -4c^2 +16ci +16 }}}
multiply, working the i^2 as well
{{{ c^4 -8c^3i -8c^2 -16c^2(-1) +16ci +16 }}}
combine like terms
{{{ c^4 - 8c^3i -8c^2 +16c^2 +16ci +16 }}}
{{{ c^4 - 8c^3i +8c^2 +16ci +16 }}}
Now using this quantity ... multiply it by the 5th (c-2i)
{{{ (c-2i)(c^4 -8c^3i +8c^2 +16ci +16) }}}
{{{ (c*c^4) + (c*-8c^3i) + (c*8c^2) + (c*16ci) + (c*16) + (-2i*c^4)+(-2i*-8c^3i)+ (-2i*8c^2) + (-2i*16ci) + (-2i*16) }}}
Multiply, working the i^2 as well
{{{ c^5 -8c^4i +8c^3 +16c^2i +16c -2c^4i +16c^3(-1) -16c^2i -32c(-1) -32i }}}
combine like terms
{{{ c^5 -10c^4i - 8c^3i + 48c - 32i }}}
the c^2 cancel out