SOLUTION: deduce the solutions of the equation (z + 1)˄4 = 16(z - 1)˄4.

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Question 1123522: deduce the solutions of the equation
(z + 1)˄4 = 16(z - 1)˄4.
Found 2 solutions by Alan3354, Edwin McCravy:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
(z + 1)^4 = 16(z - 1)^4
(z + 1)^4 = 2^4*(z - 1)^4
(z + 1)^4 = (2z - 2)^4
z+1 = 2z-2
z = 3
=================
z%5E4+%2B+4z%5E3+%2B+6z%5E2+%2B+4z+%2B+1+=+16z%5E4+-+64z%5E3+%2B+96z%5E2+-+64z+%2B+16
15z%5E4+-+68z%5E3+%2B+90z%5E2+-+68z+%2B+15+=+0
Divide by (z-3)
---
15z%5E3+-+23z%5E2+%2B+21z+-+5+=+0
By graphical methods (or by Newton's method) z = 1/3
------------
Divide by (z-1/3)
15z%5E2+-+18z+%2B+15+=+0
5z%5E2+-+6z+%2B+5+=+0
---
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 5x%5E2%2B-6x%2B5+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-6%29%5E2-4%2A5%2A5=-64.

The discriminant -64 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -64 is + or - sqrt%28+64%29+=+8.

The solution is x%5B12%5D+=+%28--6%2B-i%2Asqrt%28+-64+%29%29%2F2%5C5+=++%28--6%2B-i%2A8%29%2F2%5C5+, or
Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+5%2Ax%5E2%2B-6%2Ax%2B5+%29

z = (3/5) + 4i/5
z = (3/5) - 4i/5


Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!

Here's a better approach:

%28z+%2B+1%29%5E4+=+16%28z+-+1%29%5E4
Use the principle of square roots:
sqrt%28%28z+%2B+1%29%5E4%29+=+%22%22+%2B-+sqrt%2816%28z+-+1%29%5E4%29
%28z+%2B+1%29%5E2+=+%22%22+%2B-+4%28z+-+1%29%5E2
Break that into two equations
%28z+%2B+1%29%5E2+=+4%28z+-+1%29%5E2; %28z+%2B+1%29%5E2+=+-4%28z+-+1%29%5E2
Use the principle of square roots again:
sqrt%28%28z+%2B+1%29%5E2%29+=+%22%22+%2B-+sqrt%284%28z+-+1%29%5E2%29; sqrt%28%28z+%2B+1%29%5E2%29+=+%22%22+%2B-+sqrt%28-4%28z+-+1%29%5E2%29
z+%2B+1+=+%22%22+%2B-+2%28z+-+1%29%29; z+%2B+1+=+%22%22+%2B-+2i%28z+-+1%29%29

We solve the first using the + :
z+%2B+1+=+2%28z+-+1%29%29;
z+%2B+1+=+2z+-+2%29
-z+=+-+3%29%29
z=3

We solve the first using the - :
z+%2B+1+=+-2%28z+-+1%29%29;
z+%2B+1+=+-2z+%2B+2
3z+=1%29%29
z=1%2F3

We solve the second using the + :
z+%2B+1+=+2i%28z+-+1%29%29;
z+%2B+1+=+2iz+-+2i
z-2iz+=-1-2i
z%281-2i%29=-1-2i
z=%28-1-2i%29%2F%281-2i%29
z=%28-1-2i%29%2F%281-2i%29%22%22%2A%22%22%281%2B2i%29%2F%281%2B2i%29
z=%28-1-2i-2i-4i%5E2%29%2F%281-4i%5E2%29
z=%28-1-4i-4%28-1%29%29%2F%281-4%28-1%29%29
z=%28-1-4i%2B4%29%2F%281%2B4%29
z=%283-4i%29%2F5
z=3%2F5-expr%284%2F5%29i

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%2F5%2Bexpr%284%2F5%29i is also a solution.

The four solutions are:

3, 1%2F3, 3%2F5%2Bexpr%284%2F5%29i, and 3%2F5-expr%284%2F5%29i

Edwin