Question 1123522
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Here's a better approach:

{{{(z + 1)^4 = 16(z - 1)^4}}}
Use the principle of square roots:
{{{sqrt((z + 1)^4) = "" +- sqrt(16(z - 1)^4)}}}
{{{(z + 1)^2 = "" +- 4(z - 1)^2}}}
Break that into two equations
{{{(z + 1)^2 = 4(z - 1)^2}}}; {{{(z + 1)^2 = -4(z - 1)^2}}}
Use the principle of square roots again:
{{{sqrt((z + 1)^2) = "" +- sqrt(4(z - 1)^2)}}}; {{{sqrt((z + 1)^2) = "" +- sqrt(-4(z - 1)^2)}}}
{{{z + 1 = "" +- 2(z - 1))}}}; {{{z + 1 = "" +- 2i(z - 1))}}}

We solve the first using the + :
{{{z + 1 = 2(z - 1))}}};
{{{z + 1 = 2z - 2)}}}
{{{-z = - 3))}}}
{{{z=3}}}

We solve the first using the - :
{{{z + 1 = -2(z - 1))}}};
{{{z + 1 = -2z + 2}}}
{{{3z =1))}}}
{{{z=1/3}}}

We solve the second using the + :
{{{z + 1 = 2i(z - 1))}}};
{{{z + 1 = 2iz - 2i}}}
{{{z-2iz =-1-2i}}}
{{{z(1-2i)=-1-2i}}}
{{{z=(-1-2i)/(1-2i)}}}
{{{z=(-1-2i)/(1-2i)}}}{{{""*""}}}{{{(1+2i)/(1+2i)}}}
{{{z=(-1-2i-2i-4i^2)/(1-4i^2)}}}
{{{z=(-1-4i-4(-1))/(1-4(-1))}}}
{{{z=(-1-4i+4)/(1+4)}}}
{{{z=(3-4i)/5}}}
{{{z=3/5-expr(4/5)i}}}

We could solve the second using the - the same way,
but since we know that if a polynomial equation with real
rational coefficients has a certain complex imaginary 
solution, then its conjugate is also a solution.

So {{{z=3/5+expr(4/5)i}}} is also a solution.

The four solutions are:

{{{3}}}, {{{1/3}}}, {{{3/5+expr(4/5)i}}}, and {{{3/5-expr(4/5)i}}}

Edwin</pre></font></b>