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- Thread starter transgalactic
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the question is (z-a)^3=8

it is known that z1*z2*z3=-9

One way to approach the problem is to solve the first equation to find all possible values for (z-a). Since (z-a) is raised to the third power in this equation, we expect to find three possible values. This will give you three equations all of the form z-a=#.

I'm not completely sure if I understand what you mean by the second equation, but I'm assuming that in your notation z1, z2, and z3 are the three possible values for z. So the three equations you got earlier are z1-a=#, z2-a=#, and z3-a=# (# stands for three different numbers). I'm also assuming that a is some complex constant. (Some particular number of the form x+y*i.)

Using the three equations you got from solving the first equation and also using z1*z2*z3=-9 you can solve for z1, z2, z3, and a. This is because you now have four equations and four unknowns:

z1-a=#

z2-a=#

z3-a=#

z1*z2*z3=-9

- #3

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z1 z2 z3

i found them but transforming the equation to a trigonometric form

and then i made DEMUAVER law and found these 3 roots

then i multiplied them and i could not solve the equetion wich came out

how do i solve this thing??

- #4

HallsofIvy

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z- a= 2, or [itex]-1+ i\sqrt{3}[/itex], or [itex]-1-i\sqrt{3}[/itex].

Then z= 2+ a, or [itex]a-1+i\sqrt{3}[/itex], or [itex]a-1- i\sqrt{3}[/itex].

Multiplying those together and setting them equal to -9, you get a very

- #5

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in the end of the file i showed the equation that came out

its impossible to solve

its impossible to solve

- #6

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It's not impossible to solve. The complex numbers are algebraically closed.

- #7

HallsofIvy

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- #8

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dammmmmm

i didnt see that cube formula right under my nose

it is so simple

thank you very much

i didnt see that cube formula right under my nose

it is so simple

thank you very much

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