Question:

Step 1: To find the value of $\dfrac{12}{\pi}\arg(z_1)$, we first need to calculate $z_1$. The expression for $z_1$ is given by $z_1=\dfrac{1+i\bar{z}}{\bar{z}(1-z)+\dfrac{1}{z}}$. We know that $z=1+i$, so we will start by finding the conjugate of $z$, denoted as $\bar{z}$. Step 2: The conjugate of a complex number $z=a+bi$ is given by $\bar{z}=a-bi$. Here, $z=1+i$, so the conjugate $\bar{z}$ is $1-i$. Now, we substitute the values of $z$ and $\bar{z}$ into the expression for $z_1$ to simplify it. Step 3: Substituting $z=1+i$ and $\bar{z}=1-i$ into the expression $z_1=\dfrac{1+i\bar{z}}{\bar{z}(1-z)+\dfrac{1}{z}}$, we get: $$ z_1 = \dfrac{1+i(1-i)}{(1-i)(1-(1+i))+\dfrac{1}{1+i}} $$ Simplifying the numerator and the denominator separately will help us find $z_1$. Step 4: Simplifying the numerator $1+i(1-i)$ gives us: $$ 1+i(1-i) = 1 + i - i^2 = 1 + i + 1 = 2 + i $$ And simplifying the denominator $(1-i)(1-(1+i))+\dfrac{1}{1+i}$ gives: $$ (1-i)(-i) + \dfrac{1}{1+i} = -i + i^2 + \dfrac{1}{1+i} = -i - 1 + \dfrac{1}{1+i} $$ To add these fractions, we need a common denominator. Step 5: To simplify $\dfrac{1}{1+i}$, we multiply both the numerator and the denominator by the conjugate of the denominator, $1-i$: $$ \dfrac{1}{1+i} \cdot \dfrac{1-i}{1-i} = \dfrac{1-i}{(1+i)(1-i)} = \dfrac{1-i}{1^2-i^2} = \dfrac{1-i}{2} $$ So, the expression becomes: $$ z_1 = \dfrac{2+i}{-i-1+\dfrac{1-i}{2}} $$ Further simplification of the denominator is needed. Step 6: Simplifying the denominator $-i-1+\dfrac{1-i}{2}$ gives: $$ -i-1+\dfrac{1-i}{2} = \dfrac{-2i-2+1-i}{2} = \dfrac{-1-3i}{2} $$ Thus, $z_1$ becomes: $$ z_1 = \dfrac{2+i}{\dfrac{-1-3i}{2}} = \dfrac{2(2+i)}{-1-3i} $$ Step 7: To divide complex numbers, we multiply the numerator and the denominator by the conjugate of the denominator: $$ z_1 = \dfrac{2(2+i)}{-1-3i} \cdot \dfrac{-1+3i}{-1+3i} $$ Multiplying out the numerators and denominators gives: $$ z_1 = \dfrac{2(2+i)(-1+3i)}{(-1-3i)(-1+3i)} = \dfrac{2(-2+6i-1i+3i^2)}{1-9i^2} $$ Since $i^2 = -1$, we substitute this value in: $$ z_1 = \dfrac{2(-2+5i-3)}{1+9} = \dfrac{2(-5+5i)}{10} = \dfrac{-5+5i}{5} = -1+i $$ Step 8: Now that we have $z_1 = -1+i$, we can find $\arg(z_1)$. The argument of a complex number $z = a+bi$ is given by $\arg(z) = \tan^{-1}\left(\dfrac{b}{a}\right)$. For $z_1 = -1+i$, $a=-1$ and $b=1$, so: $$ \arg(z_1) = \tan^{-1}\left(\dfrac{1}{-1}\right) = \tan^{-1}(-1) $$ Since $\tan^{-1}(-1) = -\dfrac{\pi}{4}$, but the argument is usually given in the range $(-\pi, \pi]$, and considering the quadrant in which $z_1$ lies, we adjust the angle accordingly. Step 9: Given that $z_1$ lies in the second quadrant, where the real part is negative and the imaginary part is positive, the correct angle is $\dfrac{3\pi}{4}$ because we consider the principal argument in the range $(-\pi, \pi]$. However, the calculation directly gives us the angle in the fourth quadrant, so we need to adjust our understanding to match the given problem's context. The actual calculation of $\arg(z_1)$ from $z_1 = -1 + i$ should directly consider the quadrant, leading to $\arg(z_1) = \dfrac{3\pi}{4}$. Step 10: Finally, to find $\dfrac{12}{\pi}\arg(z_1)$, we substitute the value of $\arg(z_1)$: $$ \dfrac{12}{\pi} \cdot \dfrac{3\pi}{4} = 9 $$ Thus, the value of $\dfrac{12}{\pi}\arg(z_1)$ is $9$. The final answer is 9.