Animated Solution for Mathematics - Complex Numbers: Let $r$ and $\theta$ respectively be the modulus and amplitude of the complex number $z=2-i(2\tan \frac{5\pi }{8})$, then $(r,\theta )$ is equal to

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are peeling back the layers of a complex number to reveal its true identity.

When you look at $z=2-i(2\tan \frac{5\pi }{8})$, it might look like a jumble of trigonometry and imaginary units. Imagine this complex number as a vector pointing somewhere in the Argand plane. Our mission is to find its length (the modulus $r$) and its direction (the argument $\theta$).

To find the modulus $r$, we rely on the classic definition: $r=\sqrt{x^2+y^2}$. Here, our real part is $x=2$, and our imaginary part is $y=-2\tan \frac{5\pi }{8}$.

When we plug these into our formula, we get:

As we square these terms, the negative sign vanishes, leaving us with $r=\sqrt{4+4{\tan }^2\frac{5\pi }{8}}$. Now, factor out that $4$, and you see the beauty of the identity $1+{\tan }^2\theta ={\sec }^2\theta$ emerging from the shadows.

We are left with the following expression:

But wait! Here is where many students stumble. The square root of a square is the absolute value: $r=2|\sec \frac{5\pi }{8}|$.

Since $\frac{5\pi }{8}$ lies in the second quadrant, the secant function is negative. To make our modulus positive, we use the identity $\sec (\pi -\theta )=-\sec \theta$. Thus, $|\sec \frac{5\pi }{8}|=\sec \frac{3\pi }{8}$.

Our modulus is finally revealed as $r=2\sec \frac{3\pi }{8}$.

Now that we have the magnitude, we need the direction. We know that $\tan \theta =\frac{y}{x}$.

Let's look at our imaginary part again: $y=-2\tan \frac{5\pi }{8}$. Using the same identity as before, $\tan \frac{5\pi }{8}=\tan (\pi -\frac{3\pi }{8})=-\tan \frac{3\pi }{8}$.

Substituting this back into our expression for $z$, we get:

Look at that! Both the real part ($2$) and the imaginary part ($2\tan \frac{3\pi }{8}$) are positive. This tells us that our complex number $z$ resides firmly in the first quadrant.

This is a moment of relief—it means our argument $\theta$ is simply ${\tan }^{-1}(\frac{y}{x})$.

Calculating the argument becomes a breeze now:

We have successfully navigated the treacherous waters of the Argand plane. We found the modulus $r=2\sec \frac{3\pi }{8}$ and the argument $\theta =\frac{3\pi }{8}$.

The final polar representation $(r,\theta )$ is $(2\sec \frac{3\pi }{8},\frac{3\pi }{8})$.

Remember, the secret to these problems is never to rush. Treat each trigonometric identity as a tool in your kit, and always check your quadrant. You have the power to deconstruct even the most intimidating expressions.