The least positive integer n such that (2i)ⁿ / (1-i)ⁿ⁻², i = √-1 is a positive integer, is.

Q: Why can't we just expand $(1-i)^{n-2}$ using the binomial theorem?

While theoretically possible, the binomial theorem would lead to a massive summation of terms that are difficult to simplify into a single complex number. Using Euler's form $z = re^{i\theta}$ is the standard 'JEE-style' shortcut because it turns complex exponentiation into simple multiplication and addition of angles.

Q: Why must the phase be $2k\pi$?

A complex number $z = R e^{i\phi}$ is a positive real number if and only if its imaginary part is zero and its real part is positive. This happens when the angle $\phi$ is $0, 2\pi, 4\pi, \dots$, which is compactly written as $2k\pi$ for any integer $k$.

Q: What if $n$ is not an integer? The question asks for the least positive integer $n$.

The problem explicitly restricts $n$ to be an integer. If our derived equation $3n - 2 = 8k$ did not yield an integer $n$ for a specific $k$, we would simply test the next integer value of $k$ until we found a valid $n$.

Question

The least positive integer $n$ such that $\frac{(2i)^n}{(1-i)^{n-2}}$, $i = \sqrt{-1}$ is a positive integer, is.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the complex plane, staring at the expression $E = \frac{(2i)^n}{(1-i)^{n-2}}$. At first glance, it looks like a chaotic mess of powers and imaginary units.

As an elite JEE aspirant, you know that chaos is just order waiting to be discovered. We are not going to brute-force this with algebra; we are going to use the most powerful tool in our arsenal: **Euler's form**.

### The Euler Transformation

First, let us tame the bases. The numerator, $2i$, is a point on the positive imaginary axis. Its distance from the origin is $2$, and its angle is $\frac{\pi}{2}$. Thus, $2i = 2 e^{i\frac{\pi}{2}}$.

Now, look at the denominator, $1-i$. It sits in the fourth quadrant, one unit right and one unit down. Its magnitude is $\sqrt{1^2 + (-1)^2} = \sqrt{2}$, and its angle is $-\frac{\pi}{4}$.

So, $1-i = \sqrt{2} e^{-i\frac{\pi}{4}}$. This transformation is the key to unlocking the problem.

### The Algebraic Dance

Now, we substitute these avatars into our expression $E$:

$$E = \frac{(2 e^{i\frac{\pi}{2}})^n}{(\sqrt{2} e^{-i\frac{\pi}{4}})^{n-2}}$$

By distributing the powers, the numerator becomes $2^n e^{i\frac{n\pi}{2}}$. The denominator becomes $(2^{\frac{1}{2}})^{n-2} e^{-i\frac{(n-2)\pi}{4}}$, which simplifies to $2^{\frac{n-2}{2}} e^{-i\frac{(n-2)\pi}{4}}$.

Now, watch the magic happen as we combine the magnitudes and the phases. The magnitude part is:

$$2^{n - \frac{n-2}{2}} = 2^{\frac{n+2}{2}}$$

The phase part is $e^{i(\frac{n\pi}{2} - (-\frac{(n-2)\pi}{4}))} = e^{i(\frac{n\pi}{2} + \frac{(n-2)\pi}{4})}$. Simplifying the exponent gives us:

$$\frac{(2n\pi + n\pi - 2\pi)}{4} = \frac{(3n-2)\pi}{4}$$

We have successfully condensed the entire expression into:

$$E = 2^{\frac{n+2}{2}} e^{i\frac{(3n-2)\pi}{4}}$$

### The Geometric Constraint

For $E$ to be a positive integer, it must lie on the positive real axis. This means the phase $\frac{(3n-2)\pi}{4}$ must be a multiple of $2\pi$.

Setting $\frac{(3n-2)\pi}{4} = 2k\pi$, we cancel $\pi$ and arrive at the beautiful linear equation:

$$3n - 2 = 8k$$

We need the least positive integer $n$. Testing $k=1$ gives $3n = 10$ (no integer solution). Testing $k=2$ gives $3n = 18$, which yields **$n = 6$**.

By plugging $n=6$ back into our magnitude, we get $2^{\frac{6+2}{2}} = 2^4 = 16$, which is indeed a positive integer. You have just mastered the art of complex rotation!

Animated Solution for Mathematics - Complex Numbers: The least positive integer $n$ such that $\frac{( 2 i ) ^{n}}{( 1 - i ) ^{n - 2}}$ , $i = - 1$ is a positive integer, is.

Enter Numerical Value:

Visualized Solution

Define the Expression $E$

Euler's Form

Polar Form of $2 i$

Polar Form of $1 - i$

Substitute into Expression $E$

Expand the Numerator

Expand the Denominator

Combine the Real Magnitudes

Combine the Phases

Simplify the Total Phase

Condition for Positive Integer

Solve for Least Integer $n$

Final Conclusion

The Sigma Insight: Argand Plane and Polar Representation

Analyzing the Setup

The Euler Transformation

The Algebraic Dance

The Geometric Constraint

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Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The least positive integer n such that (1−i)n−2(2i)n​, i=−1​ is a positive integer, is.