# A complex number question

• transgalactic
In summary, the question is (z-a)^3=8. It is known that z1*z2*z3=-9. Using 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=#, and z= 2+ a, or a-1+i\sqrt{3}, or a-1- i\sqrt{3}.

#### transgalactic

i added a file in which i tried to solve this question
the final equation does"nt come out

the question is (z-a)^3=8
it is known that z1*z2*z3=-9

i have dried to make an equation

how do i solve this equation??

#### Attachments

• comp.GIF
160.7 KB · Views: 483
transgalactic said:
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

the original (z-a)^3=8 has 3 roots
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 which came out

how do i solve this thing??

I presume that when you used d'Moivre's formula you found that
z- a= 2, or $-1+ i\sqrt{3}$, or $-1-i\sqrt{3}$.
Then z= 2+ a, or $a-1+i\sqrt{3}$, or $a-1- i\sqrt{3}$.
Multiplying those together and setting them equal to -9, you get a very simple equation for a!

in the end of the file i showed the equation that came out
its impossible to solve

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

The equation you have on your file is exactly what I got: (a+2)(a2- 2a+ 4)= -9. Did you multiply it out? It is exactly a3+ 8= -9 or a3= -17. Surely you can solve that- just take the cube root.

dammmmmm
i didnt see that cube formula right under my nose
it is so simple

thank you very much

Last edited:

## 1. What is a complex number?

A complex number is a number that consists of a real part and an imaginary part. It is written in the form a + bi, where a is the real part, b is the imaginary part, and i is the imaginary unit.

## 2. How do you add or subtract complex numbers?

To add or subtract complex numbers, you simply add or subtract the real parts and the imaginary parts separately. For example, (2 + 3i) + (4 + 5i) = (2 + 4) + (3 + 5)i = 6 + 8i. Similarly, (2 + 3i) - (4 + 5i) = (2 -4) + (3 - 5)i = -2 - 2i.

## 3. What is the conjugate of a complex number?

The conjugate of a complex number is a number with the same real part but opposite imaginary part. For example, the conjugate of 2 + 3i is 2 - 3i.

## 4. How do you multiply complex numbers?

To multiply complex numbers, you can use the FOIL method, just like multiplying binomials. For example, (2 + 3i)(4 + 5i) = 2(4) + 2(5i) + 3i(4) + 3i(5i) = 8 + 10i + 12i + 15i^2 = 8 + 22i - 15 = -7 + 22i.

## 5. What is the complex conjugate of a complex number used for?

The complex conjugate is useful in dividing complex numbers and simplifying complex fractions. When dividing complex numbers, you can multiply the numerator and denominator by the complex conjugate of the denominator to eliminate the imaginary part in the denominator. This makes the division easier to solve.