Let \( z \) be a complex number of unit modulus, which means that \[ |z| = 1. \] We need to evaluate: \[ \left| \frac{1 + z}{1 + \overline{z}} \right|. \] Step 1: Express \( \overline{z} \) in terms of \( z \) The complex conjugate of \( z \) is denoted by \( \overline{z} \), and for any complex number \( z = x + iy \), its conjugate is given by \( \overline{z} = x - iy \). Since \( z \) has unit modulus, we know that:
\[ |z|^2 = 1 \Rightarrow z \overline{z} = 1. \] Step 2: Simplify the expression
We can simplify the expression as follows: \[ \left| \frac{1+z}{1+\overline{z}} \right| = \frac{|1+z|}{|1+\overline{z}|}. \] Step 3: Evaluate \( |1+z| \) and \( |1+\overline{z}| \) Using the fact that \( |z| = 1 \), we compute both the modulus of \( 1+z \) and \( 1+\overline{z} \).
Since the modulus of a complex number is the distance from the origin, and \( |z| = 1 \), the expressions for \( |1+z| \) and \( |1+\overline{z}| \) are equal.
Thus: \[ \left| \frac{1+z}{1+\overline{z}} \right| = 1. \]
Therefore, the correct answer is option (B), which is 1.
Let \( f(x) = \frac{x^2 + 40}{7x} \), \( x \neq 0 \), \( x \in [4,5] \). The value of \( c \) in \( [4,5] \) at which \( f'(c) = -\frac{1}{7} \) is equal to:
The general solution of the differential equation \( \frac{dy}{dx} = xy - 2x - 2y + 4 \) is:
The minimum value of the function \( f(x) = x^4 - 4x - 5 \), where \( x \in \mathbb{R} \), is:
The critical points of the function \( f(x) = (x-3)^3(x+2)^2 \) are:
\[ f(x) = \begin{cases} x\left( \frac{\pi}{2} + x \right), & \text{if } x \geq 0 \\ x\left( \frac{\pi}{2} - x \right), & \text{if } x < 0 \end{cases} \]
Then \( f'(-4) \) is equal to:If \( f'(x) = 4x\cos^2(x) \sin\left(\frac{x}{4}\right) \), then \( \lim_{x \to 0} \frac{f(\pi + x) - f(\pi)}{x} \) is equal to:
\[ \int \frac{4x \cos \left( \sqrt{4x^2 + 7} \right)}{\sqrt{4x^2 + 7}} \, dx \]
\[ \int \left( \frac{\log_e t}{1+t} + \frac{\log_e t}{t(1+t)} \right) dt \]