Animated Solution for Mathematics - Complex Numbers: Let $z=\cos \theta +i\sin \theta$. Then the value of ${\sum }_{m=1}^{15}Im(z^{2m-1})$ at $\theta =2^{\circ }$ is

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are witnessing the beautiful harmony between complex numbers and trigonometry.

When you see $z=\cos \theta +i\sin \theta$, do not just see a static expression. See a rotation operator. By Euler's formula, we know that $z=e^{i\theta }$.

Geometrically, this is a point on the unit circle in the complex plane, making an angle $\theta$ with the positive real axis. When we raise this to a power, we are simply scaling the angle.

Specifically, $z^{2m-1}=e^{i(2m-1)\theta }=\cos ((2m-1)\theta )+i\sin ((2m-1)\theta )$. The imaginary part, which we are tasked to sum, is simply the vertical projection of this point: $\sin ((2m-1)\theta )$.

Now, let us look at the summation: $S={\sum }_{m=1}^{15}\sin ((2m-1)\theta )$. If we expand this, we get $\sin (\theta )+\sin (3\theta )+\sin (5\theta )+\cdots +\sin (29\theta )$.

Do you see the rhythm? The angles are $\theta ,3\theta ,5\theta ,\dots ,29\theta$. This is a classic Arithmetic Progression (A.P.) with the first term $\alpha =\theta$, the common difference $d=2\theta$, and the number of terms $n=15$.

Recognizing this structure is the key to unlocking the problem without brute force.

In the heat of the exam, you need tools that are sharp and reliable. For a sum of sines where the angles are in an A.P., we have a powerful identity:

This formula is your best friend. Let's substitute our values: $n=15$, $\alpha =\theta$, and $d=2\theta$. The expression becomes:

Simplifying this is where the magic happens. The first fraction simplifies to $\frac{\sin (15\theta )}{\sin (\theta )}$. The second sine term simplifies to $\sin (\theta +14\theta )=\sin (15\theta )$.

Look at what we have created:

This is the compact, elegant soul of the problem. Now, we simply apply the given condition $\theta =2^{\circ }$. Thus, $15\theta =30^{\circ }$.

We know that $\sin (30^{\circ })=\frac{1}{2}$, so ${\sin }^2(30^{\circ })=\frac{1}{4}$. Substituting this back, we get:

And there it is! The complexity has vanished, leaving behind a simple, beautiful result that matches option 4. Remember, in physics and mathematics, complexity is often just a mask for a deeper, simpler truth. Keep looking for that truth, and you will never be lost.