if ` z = cos ""pi/4 + i sin""pi/6` then

Options

A. `|z|=1 , arg(z)= pi/4`
B. `|z|=1, arg(z)= pi/6`
C. `|z|=sqrt3/2, arg(z)= (5pi)/24`
D. `|z| = sqrt3/2 , arg(z)= tan^(-1)1/sqrt2`

Answer

To solve the problem given \( z = \cos \frac{\pi}{4} + i \sin \frac{\pi}{6} \), we will follow these steps:

### Step 1: Calculate \( \cos \frac{\pi}{4} \) and \( \sin \frac{\pi}{6} \)

Using the known values of trigonometric functions:
- \( \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \)
- \( \sin \frac{\pi}{6} = \frac{1}{2} \)

So, we can write:
\[ z = \frac{1}{\sqrt{2}} + i \cdot \frac{1}{2} \]

### Step 2: Find the modulus of \( z \)

The modulus of a complex number \( z = a + ib \) is given by:
\[ |z| = \sqrt{a^2 + b^2} \]
Here, \( a = \frac{1}{\sqrt{2}} \) and \( b = \frac{1}{2} \).

Calculating \( a^2 \) and \( b^2 \):
\[ a^2 = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]
\[ b^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

Now, substituting these values into the modulus formula:
\[ |z| = \sqrt{\frac{1}{2} + \frac{1}{4}} = \sqrt{\frac{2}{4} + \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

### Step 3: Find the argument of \( z \)

The argument of a complex number \( z = a + ib \) can be found using:
\[ \text{arg}(z) = \tan^{-1}\left(\frac{b}{a}\right) \]
Substituting the values of \( a \) and \( b \):
\[ \text{arg}(z) = \tan^{-1}\left(\frac{\frac{1}{2}}{\frac{1}{\sqrt{2}}}\right) = \tan^{-1}\left(\frac{1}{2} \cdot \sqrt{2}\right) = \tan^{-1}\left(\frac{\sqrt{2}}{2}\right) \]

### Final Result

Thus, we have:
- The modulus of \( z \) is \( |z| = \frac{\sqrt{3}}{2} \)
- The argument of \( z \) is \( \text{arg}(z) = \tan^{-1}\left(\frac{\sqrt{2}}{2}\right) \)