Read the following writeup carefully:
If `z_1 = a+ib and z_2 =c + id` be two complex numbers such that `|z_1| = |z_2|=1 and "Re" (z_1 bar(z_2))=0`.
Now answer the following question
Let `k = |z_1 + z_2| + |a+ ib|`, then the value of k is

Options

A. `sqrt2 -1`
B. 1
C. `sqrt2+1`
D. ` 2 sqrt2`

Answer

To solve the given problem step by step, let's break it down as follows:

### Step 1: Understand the given complex numbers
Let \( z_1 = a + ib \) and \( z_2 = c + id \). We know that \( |z_1| = 1 \) and \( |z_2| = 1 \). This implies:
\[ |z_1| = \sqrt{a^2 + b^2} = 1 \quad \text{(1)} \]
\[ |z_2| = \sqrt{c^2 + d^2} = 1 \quad \text{(2)} \]

### Step 2: Use the modulus conditions
From equations (1) and (2), we can square both sides:
\[ a^2 + b^2 = 1 \quad \text{(3)} \]
\[ c^2 + d^2 = 1 \quad \text{(4)} \]

### Step 3: Analyze the condition on the real part
We are given that \( \text{Re}(z_1 \overline{z_2}) = 0 \). The conjugate of \( z_2 \) is \( \overline{z_2} = c - id \). Therefore:
\[ z_1 \overline{z_2} = (a + ib)(c - id) = ac + bd + i(bc - ad) \]
The real part is:
\[ \text{Re}(z_1 \overline{z_2}) = ac + bd = 0 \quad \text{(5)} \]

### Step 4: Calculate \( k = |z_1 + z_2| + |z_1| \)
We know \( |z_1| = 1 \), so we need to find \( |z_1 + z_2| \):
\[ z_1 + z_2 = (a + c) + i(b + d) \]
Thus,
\[ |z_1 + z_2| = \sqrt{(a + c)^2 + (b + d)^2} \]

### Step 5: Expand \( |z_1 + z_2|^2 \)
Now, we can expand \( |z_1 + z_2|^2 \):
\[ |z_1 + z_2|^2 = (a + c)^2 + (b + d)^2 \]
Expanding this gives:
\[ = a^2 + 2ac + c^2 + b^2 + 2bd + d^2 \]
Using equations (3) and (4):
\[ = (a^2 + b^2) + (c^2 + d^2) + 2(ac + bd) = 1 + 1 + 2(0) = 2 \]
Thus,
\[ |z_1 + z_2|^2 = 2 \implies |z_1 + z_2| = \sqrt{2} \]

### Step 6: Calculate \( k \)
Now substituting back into \( k \):
\[ k = |z_1 + z_2| + |z_1| = \sqrt{2} + 1 \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{\sqrt{2} + 1} \]