Animated Solution for Mathematics - Complex Numbers: Let $z=1+i$ and $z_1=\frac{1+i\bar{z}}{\bar{z}(1-z)+\frac{1}{z}}$. Then $\frac{12}{\pi }\arg (z_1)$ is equal to \_\_\_\_\_.

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving an equation; we are exploring the geometric elegance of complex numbers.

When you look at an expression like $z_1=\frac{1+i\bar{z}}{\bar{z}(1-z)+\frac{1}{z}}$, it is easy to feel overwhelmed. But remember, every complex expression is just a dance of rotations and scalings. Let us break this down, step by step, with the precision of a surgeon and the curiosity of a mathematician.

The Building Blocks

We begin with $z=1+i$. On the Argand plane, this point sits comfortably in the first quadrant, at coordinates $(1,1)$.

Its conjugate, $\bar{z}=1-i$, is its reflection across the real axis. These two values are the keys to our kingdom. Whenever you see $z$ and $\bar{z}$ in an expression, keep them close; they are the foundation upon which we will build our solution.

The Numerator's Dance

Let us isolate the numerator: $1+i\bar{z}$. Substituting $\bar{z}=1-i$, we get $1+i(1-i)$.

Distributing the $i$, we have $1+i-i^2$. Here is where the magic happens: since $i^2=-1$, our expression becomes $1+i-(-1)$, which simplifies beautifully to $2+i$.

See how the complexity melts away when we handle it with care?

The Denominator's Complexity

Now, let us tackle the denominator, which is the sum of two parts: $\bar{z}(1-z)$ and $\frac{1}{z}$.

First, we calculate the product:

Second, we rationalize the reciprocal:

Combining these, the denominator is $(-1-i)+\frac{1-i}{2}$. Taking a common denominator of $2$, we obtain:

The Grand Unification

Now we bring it all together. That $2$ in the denominator's denominator flips up to the numerator:

To reach the standard form, we rationalize by multiplying the numerator and denominator by the conjugate of the denominator, which is $-1+3i$:

Simplifying this, we find $z_1=-1+i$.

The Final Argument

We have arrived at $z_1=-1+i$. Plotting this, we see it lies in the second quadrant.

The reference angle is $\alpha ={\tan }^{-1}\left|\frac{1}{-1}\right|=\frac{\pi }{4}$. Since it is in the second quadrant, the argument is:

Finally, the question asks for the value of $\frac{12}{\pi }\arg (z_1)$:

The elegance of the result—a simple integer—is the reward for our persistence. You have navigated the complexity and emerged victorious. Keep this confidence for your next challenge!