Which of the following represents a points in an Argand pane, equidistant from the roots of the equation `(z+1)^4=16 z^4?` `(0,0)` b. `(-1/3,0)` c. `(1/3,0)` d. `(0,2/(sqrt(5)))`

Options

A. `(0,0)`
B. `(-(1)/(3),0)`
C. `((1)/(3),0)`
D. `(0,(2)/(sqrt(5)))`

Explanation

To find the point in the Argand plane that is equidistant from the roots of the equation \((z+1)^4 = 16z^4\), we can follow these steps:

### Step 1: Rewrite the Equation
Start by rewriting the equation:
\[ (z+1)^4 = 16z^4 \]
Divide both sides by \(z^4\) (assuming \(z \neq 0\)):
\[ \left(\frac{z+1}{z}\right)^4 = 16 \]

### Step 2: Take the Fourth Root
Taking the fourth root of both sides gives us:
\[ \frac{z+1}{z} = \pm 2, \pm 2i \]

### Step 3: Solve for \(z\)
Now, we can solve for \(z\) in each case.

1. For \(\frac{z+1}{z} = 2\):
\[ z + 1 = 2z \implies z = 1 \]

2. For \(\frac{z+1}{z} = -2\):
\[ z + 1 = -2z \implies 3z = -1 \implies z = -\frac{1}{3} \]

3. For \(\frac{z+1}{z} = 2i\):
\[ z + 1 = 2iz \implies z - 2iz = -1 \implies z(1 - 2i) = -1 \implies z = \frac{-1}{1 - 2i} \]
To simplify, multiply the numerator and denominator by the conjugate:
\[ z = \frac{-1(1 + 2i)}{(1 - 2i)(1 + 2i)} = \frac{-1 - 2i}{1 + 4} = \frac{-1 - 2i}{5} = -\frac{1}{5} - \frac{2}{5}i \]

4. For \(\frac{z+1}{z} = -2i\):
\[ z + 1 = -2iz \implies z + 2iz = -1 \implies z(1 + 2i) = -1 \implies z = \frac{-1}{1 + 2i} \]
Again, multiply by the conjugate:
\[ z = \frac{-1(1 - 2i)}{(1 + 2i)(1 - 2i)} = \frac{-1 + 2i}{1 + 4} = \frac{-1 + 2i}{5} = -\frac{1}{5} + \frac{2}{5}i \]

### Step 4: List the Roots
The roots we found are:
1. \(z = 1\)
2. \(z = -\frac{1}{3}\)
3. \(z = -\frac{1}{5} - \frac{2}{5}i\)
4. \(z = -\frac{1}{5} + \frac{2}{5}i\)

### Step 5: Find the Midpoint
To find a point that is equidistant from these roots, we can look for the midpoint between the real roots \(z = 1\) and \(z = -\frac{1}{3}\):
\[ \text{Midpoint} = \left(\frac{1 + (-\frac{1}{3})}{2}, 0\right) = \left(\frac{2/3}{2}, 0\right) = \left(\frac{1}{3}, 0\right) \]

### Conclusion
Thus, the point that is equidistant from the roots of the equation is:
\[ \boxed{\left(\frac{1}{3}, 0\right)} \]