Question

Let a, b, c be distinct complex numbers such that $\frac{a}{1-b}=\frac{b}{1-c}=\frac{c}{1-a}=k$.

Find the value of “k”.

Answer 1

$\frac{1 \pm i\sqrt{3}}{2}$

Answer 2

Let the given equations be:

1. $a = k(1-b)$
2. $b = k(1-c)$
3. $c = k(1-a)$

We are given that $a, b, c$ are distinct complex numbers.

Step 1: Check for simple values of k.

• If $k=0$, then $a=0, b=0, c=0$. This contradicts the condition that $a, b, c$ are distinct. So $k \neq 0$.

• If $k=1$, then:

  $a = 1-b \implies a+b=1$ $b = 1-c \implies b+c=1$ $c = 1-a \implies c+a=1$

  From $a+b=1$ and $b+c=1$, we get $a=c$. This contradicts the condition that $a, b, c$ are distinct. So $k \neq 1$.

• If $k=-1$, then:

  $a = -(1-b) = b-1$ $b = -(1-c) = c-1$ $c = -(1-a) = a-1$

  Substitute $b=a+1$ into the second equation: $a+1 = c-1 \implies c = a+2$.

  Substitute $c=a+2$ into the third equation: $a+2 = a-1 \implies 2=-1$. This is a contradiction. So $k \neq -1$.

Step 2: Substitute cyclically to find a relationship for 'a'.

From (1), $a = k - kb$.

Substitute $b = k(1-c)$ into this:

$a = k - k[k(1-c)] = k - k^2(1-c) = k - k^2 + k^2c$.

Substitute $c = k(1-a)$ into this:

$a = k - k^2 + k^2[k(1-a)] = k - k^2 + k^3(1-a) = k - k^2 + k^3 - k^3a$.

Rearrange the terms to solve for $a$:

$a + k^3a = k - k^2 + k^3$

$a(1+k^3) = k - k^2 + k^3$

Similarly, we would find:

$b(1+k^3) = k - k^2 + k^3$

$c(1+k^3) = k - k^2 + k^3$

Step 3: Use the distinctness condition.

If $1+k^3 \neq 0$, then $a = b = c = \frac{k - k^2 + k^3}{1+k^3}$. This contradicts the condition that $a, b, c$ are distinct.

Therefore, it must be that $1+k^3 = 0$.

If $1+k^3 = 0$, then the equation $a(1+k^3) = k - k^2 + k^3$ becomes $a(0) = k - k^2 + k^3$.

For a solution to exist, the right-hand side must also be zero:

$k - k^2 + k^3 = 0$.

So we need both conditions to be true:

(i) $1+k^3 = 0 \implies k^3 = -1$

(ii) $k - k^2 + k^3 = 0$

Substitute $k^3=-1$ into (ii):

$k - k^2 - 1 = 0$

$k^2 - k + 1 = 0$

Step 4: Solve the quadratic equation for k.

The roots of $k^2 - k + 1 = 0$ are given by the quadratic formula:

$k = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(1)}}{2(1)}$

$k = \frac{1 \pm \sqrt{1-4}}{2}$

$k = \frac{1 \pm \sqrt{-3}}{2}$

$k = \frac{1 \pm i\sqrt{3}}{2}$

Let's check if these values satisfy $k^3=-1$.

If $k = \frac{1+i\sqrt{3}}{2}$, then $k = e^{i\pi/3}$.

$k^3 = (e^{i\pi/3})^3 = e^{i\pi} = -1$.

If $k = \frac{1-i\sqrt{3}}{2}$, then $k = e^{-i\pi/3}$.

$k^3 = (e^{-i\pi/3})^3 = e^{-i\pi} = -1$.

Both values satisfy $k^3=-1$.

Also, we already established $k \neq -1$, and these two values are not $-1$.

Step 5: Verify that distinct a, b, c exist for these k values.

If $k^2-k+1=0$, then $k^3=-1$. The equation $a(0)=0$ holds for any $a$.

Let's choose a specific value for $a$, for example $a=1$.

If $k = \frac{1+i\sqrt{3}}{2}$:

$a=1$

$b = k(1-c)$

$c = k(1-a) = k(1-1) = 0$

Now, substitute $c=0$ into the second equation:

$b = k(1-0) = k$.

So $b = \frac{1+i\sqrt{3}}{2}$.

Now check the first equation: $a = k(1-b)$.

$1 = \frac{1+i\sqrt{3}}{2} \left(1 - \frac{1+i\sqrt{3}}{2}\right)$

$1 = \frac{1+i\sqrt{3}}{2} \left(\frac{2 - 1 - i\sqrt{3}}{2}\right)$

$1 = \frac{1+i\sqrt{3}}{2} \left(\frac{1 - i\sqrt{3}}{2}\right)$

$1 = \frac{1^2 - (i\sqrt{3})^2}{4} = \frac{1 - (-3)}{4} = \frac{1+3}{4} = \frac{4}{4} = 1$.

This is consistent.

So for $k=\frac{1+i\sqrt{3}}{2}$, we have $a=1, b=\frac{1+i\sqrt{3}}{2}, c=0$. These are distinct.

If $k = \frac{1-i\sqrt{3}}{2}$:

Similarly, if $a=1$, then $c=0$, and $b=k=\frac{1-i\sqrt{3}}{2}$. These are also distinct.

Thus, the values of $k$ are $\frac{1+i\sqrt{3}}{2}$ and $\frac{1-i\sqrt{3}}{2}$.

The final answer is $\boxed{\frac{1 \pm i\sqrt{3}}{2}}$.