If `x=9^(1/3) 9^(1/9) 9^(1/27) ......ad` inf `y= 4^(1/3) 4^(-1/9) 4^(1/27) ...... ad` inf and `z= sum_(r=1)^oo (1+i)^-r` then , the argument of the complex number `w = x+yz` is

Options

A. 0
B. `-tan^(-1)(sqrt(2)/(3))`
C. `-tan^(-1)((2)/sqrt(3))`
D. `pi-tan^(-1)(sqrt(2)/(3))`

Answer

To solve the problem step by step, we need to find the values of \( x \), \( y \), and \( z \), and then compute \( w = x + yz \) to find the argument of \( w \).

### Step 1: Calculate \( x \)

Given:
\[ x = 9^{1/3} \cdot 9^{1/9} \cdot 9^{1/27} \cdots \]

This can be rewritten by adding the exponents:
\[ x = 9^{(1/3 + 1/9 + 1/27 + \cdots)} \]

The series \( 1/3 + 1/9 + 1/27 + \cdots \) is a geometric series where:
- First term \( a = \frac{1}{3} \)
- Common ratio \( r = \frac{1}{3} \)

The sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2} \]

Thus,
\[ x = 9^{1/2} = 3 \]

### Step 2: Calculate \( y \)

Given:
\[ y = 4^{1/3} \cdot 4^{-1/9} \cdot 4^{1/27} \cdots \]

This can also be rewritten by adding the exponents:
\[ y = 4^{(1/3 - 1/9 + 1/27 + \cdots)} \]

The series \( 1/3 - 1/9 + 1/27 + \cdots \) is a geometric series where:
- First term \( a = \frac{1}{3} \)
- Common ratio \( r = -\frac{1}{3} \)

The sum of this series is:
\[ S = \frac{a}{1 - r} = \frac{\frac{1}{3}}{1 - (-\frac{1}{3})} = \frac{\frac{1}{3}}{\frac{4}{3}} = \frac{1}{4} \]

Thus,
\[ y = 4^{1/4} = \sqrt{2} \]

### Step 3: Calculate \( z \)

Given:
\[ z = \sum_{r=1}^{\infty} (1+i)^{-r} \]

This is a geometric series where:
- First term \( a = (1+i)^{-1} \)
- Common ratio \( r = (1+i)^{-1} \)

The sum is:
\[ z = \frac{a}{1 - r} = \frac{(1+i)^{-1}}{1 - (1+i)^{-1}} = \frac{(1+i)^{-1}}{\frac{(1+i)-1}{1+i}} = \frac{(1+i)^{-1} \cdot (1+i)}{i} = \frac{1}{i} = -i \]

### Step 4: Calculate \( w \)

Now we can find \( w \):
\[ w = x + yz = 3 + \sqrt{2}(-i) = 3 - \sqrt{2}i \]

### Step 5: Find the argument of \( w \)

The argument of a complex number \( a + bi \) is given by:
\[ \text{arg}(w) = \tan^{-1}\left(\frac{b}{a}\right) \]
where \( a = 3 \) and \( b = -\sqrt{2} \).

Thus,
\[ \text{arg}(w) = \tan^{-1}\left(\frac{-\sqrt{2}}{3}\right) \]

### Final Answer

The argument of the complex number \( w \) is:
\[ \text{arg}(w) = \tan^{-1}\left(-\frac{\sqrt{2}}{3}\right) \]