Number of solutions of `|z|=10andarg((z)/(3)+(i)/(3))=(pi)/(3)` is :

Answer

To solve the problem of finding the number of solutions for the equations \( |z| = 10 \) and \( \arg\left(\frac{z}{3} + \frac{i}{3}\right) = \frac{\pi}{3} \), we can follow these steps:

### Step 1: Understanding the Modulus Condition
The first condition is \( |z| = 10 \). This means that the complex number \( z \) lies on a circle of radius 10 centered at the origin in the complex plane.

### Step 2: Expressing \( z \) in Terms of Real and Imaginary Parts
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. The modulus condition can be expressed as:
\[ |z| = \sqrt{x^2 + y^2} = 10 \]
Squaring both sides gives:
\[ x^2 + y^2 = 100 \quad \text{(Equation 1)} \]

### Step 3: Understanding the Argument Condition
The second condition is \( \arg\left(\frac{z}{3} + \frac{i}{3}\right) = \frac{\pi}{3} \). We can rewrite this as:
\[ \arg\left(\frac{x + iy}{3} + \frac{i}{3}\right) = \frac{\pi}{3} \]
This simplifies to:
\[ \arg\left(\frac{x}{3} + \frac{y + 1}{3}i\right) = \frac{\pi}{3} \]
The argument of a complex number \( a + bi \) is given by \( \tan^{-1}\left(\frac{b}{a}\right) \). Therefore, we have:
\[ \tan^{-1}\left(\frac{y + 1}{x}\right) = \frac{\pi}{3} \]

### Step 4: Finding the Tangent
Taking the tangent of both sides gives:
\[ \frac{y + 1}{x} = \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \]
This leads to:
\[ y + 1 = \sqrt{3}x \quad \text{(Equation 2)} \]
Rearranging gives:
\[ y = \sqrt{3}x - 1 \]

### Step 5: Substituting Equation 2 into Equation 1
Now we substitute Equation 2 into Equation 1:
\[ x^2 + (\sqrt{3}x - 1)^2 = 100 \]
Expanding the second term:
\[ x^2 + (3x^2 - 2\sqrt{3}x + 1) = 100 \]
Combining like terms gives:
\[ 4x^2 - 2\sqrt{3}x + 1 - 100 = 0 \]
This simplifies to:
\[ 4x^2 - 2\sqrt{3}x - 99 = 0 \]

### Step 6: Solving the Quadratic Equation
We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 4 \), \( b = -2\sqrt{3} \), and \( c = -99 \):
\[ x = \frac{2\sqrt{3} \pm \sqrt{(-2\sqrt{3})^2 - 4 \cdot 4 \cdot (-99)}}{2 \cdot 4} \]
Calculating the discriminant:
\[ (-2\sqrt{3})^2 = 12 \]
\[ 4 \cdot 4 \cdot 99 = 1584 \]
So the discriminant is:
\[ 12 + 1584 = 1596 \]
Now substituting back:
\[ x = \frac{2\sqrt{3} \pm \sqrt{1596}}{8} \]

### Step 7: Finding the Number of Solutions
The quadratic equation will yield two real solutions for \( x \) (since the discriminant is positive). For each \( x \), there is a corresponding \( y \) from Equation 2. Therefore, there are two pairs \((x, y)\) that satisfy both conditions.

### Conclusion
Thus, the number of solutions for \( z \) is:
\[ \boxed{2} \]