Animated Solution for Mathematics - Complex Numbers: The value of ${\left(\frac{-1+i\sqrt{3}}{1-i}\right)}^{30}$ is:

Welcome, future engineers. Today, we are looking at a problem that serves as a perfect litmus test for your mathematical maturity.

When you see an expression like ${\left(\frac{-1+i\sqrt{3}}{1-i}\right)}^{30}$, your first instinct might be to panic or reach for a binomial expansion. But stop. Take a breath.

In JEE Advanced, brute force is rarely the intended path. The beauty of complex numbers lies in their geometry, not just their algebra. We are going to solve this by stepping into the world of Euler's form, where powers become simple multiplication.

The Geometry of the Numerator and Denominator

Let us define our components: $z_1=-1+i\sqrt{3}$ and $z_2=1-i$.

If you plot $z_1$ on the Argand plane, you see it sits in the second quadrant. Its modulus is $|z_1|=\sqrt{(-1{)}^2+(\sqrt{3}{)}^2}=2$.

Its argument, ${\theta }_1$, is $\pi -{\tan }^{-1}(\sqrt{3})=\frac{2\pi }{3}$. Thus, we can write:

Now, look at $z_2=1-i$. This vector points into the fourth quadrant. Its modulus is $|z_2|=\sqrt{1^2+(-1{)}^2}=\sqrt{2}$.

Its argument, ${\theta }_2$, is $-{\tan }^{-1}(1)=-\frac{\pi }{4}$. So, we have:

The Elegance of Division

Now, watch what happens when we divide. We are not dealing with messy conjugates anymore; we are dealing with exponents.

The ratio is:

The moduli divide simply: $\frac{2}{\sqrt{2}}=\sqrt{2}$. The exponents subtract: $\frac{2\pi }{3}-(-\frac{\pi }{4})=\frac{8\pi +3\pi }{12}=\frac{11\pi }{12}$.

Our ratio is now a clean, manageable $\sqrt{2}{e}^{i\frac{11\pi }{12}}$.

The Power of 30

Now we apply the power of 30. This is where the magic happens.

We raise our result to the 30th power:

The modulus part becomes $(2^{1/2}{)}^{30}=2^{15}$. The exponential part becomes $e^{i\frac{11\pi \times 30}{12}}$, which simplifies to $e^{i\frac{55\pi }{2}}$.

The Final Simplification

We are left with $2^{15}{e}^{i\frac{55\pi }{2}}$. We need to interpret $\frac{55\pi }{2}$.

We can write this as $27\pi +\frac{\pi }{2}$. Since $27\pi$ is an odd multiple of $\pi$, $e^{i27\pi }=-1$.

And $e^{i\frac{\pi }{2}}$ is simply $i$. Therefore, $e^{i\frac{55\pi }{2}}=(-1)(i)=-i$.

Our final answer is $-2^{15}i$.

See how we avoided the nightmare of expansion? By respecting the geometry of complex numbers, we turned a terrifying problem into a series of elegant, logical steps. Keep this mindset, and you will conquer any problem the exam throws at you.