Statement-1: for any non-zero complex number `|z-1| le ||z|-1|+|z|` arg (z)
Statement-2 : For any non-zero complex number z
`|z/(|z|)-1| le "arg"(z)`

Options

A. Statement-1 is True, Statement-2 is True: Statement-2 is a correct exp,anation for statement-1.
B. Statement-1 is true, statement -2 is true, Statement-2 is not a correct explanation for statement-1.
C. Statement-1 is True, statement-2 is false,
D. statement-1 is False, Statement-2 is true.

Answer

To solve the given problem, we need to analyze both statements and show that they are true. Let's break down the solution step by step.

### Step 1: Understanding Statement 2
The second statement is:
\[ \left| \frac{z}{|z|} - 1 \right| \leq \arg(z) \]
where \( z \) is a non-zero complex number.

**Hint for Step 1:** Recall that for any complex number \( z = r(\cos \theta + i \sin \theta) \), where \( r = |z| \) and \( \theta = \arg(z) \).

### Step 2: Expressing \( z \)
We can express \( z \) in terms of its modulus and argument:
\[ z = r(\cos \theta + i \sin \theta) \]
Thus, we have:
\[ |z| = r \]
and
\[ \arg(z) = \theta \]

**Hint for Step 2:** Remember that the modulus \( |z| \) is the distance from the origin in the complex plane.

### Step 3: Simplifying the Left-Hand Side of Statement 2
Now, we compute the left-hand side:
\[ \frac{z}{|z|} = \frac{r(\cos \theta + i \sin \theta)}{r} = \cos \theta + i \sin \theta \]
Thus,
\[ \left| \frac{z}{|z|} - 1 \right| = \left| (\cos \theta + i \sin \theta) - 1 \right| = \left| (\cos \theta - 1) + i \sin \theta \right| \]

**Hint for Step 3:** Use the formula for the modulus of a complex number: \( |a + bi| = \sqrt{a^2 + b^2} \).

### Step 4: Calculating the Modulus
Now we calculate:
\[ \left| (\cos \theta - 1) + i \sin \theta \right| = \sqrt{(\cos \theta - 1)^2 + \sin^2 \theta} \]
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ = \sqrt{(\cos^2 \theta - 2\cos \theta + 1) + \sin^2 \theta} = \sqrt{2 - 2\cos \theta} = \sqrt{2(1 - \cos \theta)} = \sqrt{2 \cdot 2\sin^2\left(\frac{\theta}{2}\right)} = 2\left|\sin\left(\frac{\theta}{2}\right)\right| \]

**Hint for Step 4:** Use the half-angle identity: \( 1 - \cos \theta = 2\sin^2\left(\frac{\theta}{2}\right) \).

### Step 5: Establishing the Inequality
Now we need to show:
\[ 2\left|\sin\left(\frac{\theta}{2}\right)\right| \leq \theta \]
This is true since \( \sin x \leq x \) for \( x \geq 0 \).

**Hint for Step 5:** Recall the property of the sine function that \( \sin x \leq x \) for all \( x \geq 0 \).

### Step 6: Conclusion for Statement 2
Thus, we conclude that Statement 2 is true.

### Step 7: Understanding Statement 1
The first statement is:
\[ |z - 1| \leq ||z| - 1| + |z| \cdot \arg(z) \]

**Hint for Step 7:** Use the triangle inequality to analyze the modulus expressions.

### Step 8: Applying the Triangle Inequality
We can rewrite \( z \) as:
\[ z = |z|(\cos \theta + i \sin \theta) \]
Then,
\[ |z - 1| = | |z|(\cos \theta + i \sin \theta) - 1 | \]

**Hint for Step 8:** Remember that the triangle inequality states \( |a + b| \leq |a| + |b| \).

### Step 9: Breaking Down the Expression
Using the triangle inequality:
\[ |z - 1| = ||z|(\cos \theta + i \sin \theta) - 1| \leq ||z| - 1| + |z| \]

**Hint for Step 9:** Consider \( |z| \) as a constant and analyze how it interacts with the other terms.

### Step 10: Finalizing Statement 1
Since we have established that both statements are true, we conclude that:
- Statement 1 is true.
- Statement 2 is true.

Thus, both statements are validated.

### Final Answer
Both Statement 1 and Statement 2 are true.