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Let's express the complex number \( 5 - 5i \) in trigonometric form.

The general form of a complex number in trigonometric (polar) form is:

\[
r(\cos \theta + i \sin \theta)
\]

Where:
- \( r \) is the modulus of the complex number.
- \( \theta \) is the argument (angle) of the complex number.

### Step 1: Find the modulus \( r \)
The modulus \( r \) is given by:

\[
r = \sqrt{a^2 + b^2}
\]

where \( a \) is the real part and \( b \) is the imaginary part of the complex number \( z = a + bi \).

For \( 5 - 5i \):
- \( a = 5 \)
- \( b = -5 \)

\[
r = \sqrt{5^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}
\]

### Step 2: Find the argument \( \theta \)
The argument \( \theta \) is the angle that the complex number makes with the positive real axis, which can be calculated as:

\[
\theta = \tan^{-1}\left(\frac{b}{a}\right)
\]

For \( 5 - 5i \):
\[
\theta = \tan^{-1}\left(\frac{-5}{5}\right) = \tan^{-1}(-1)
\]

The value of \( \tan^{-1}(-1) \) is \( 315^\circ \) (since the complex number is in the fourth quadrant, where both real part is positive and imaginary part is negative).

### Final Trigonometric Form
Now, we can write the complex number in trigonometric form:

\[
5 - 5i = 5\sqrt{2} \left( \cos 315^\circ + i \sin 315^\circ \right)
\]

Let me know if you need further clarification or a breakdown of any specific steps.

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**Relative Questions:**
1. How do you find the modulus of a complex number?
2. What is the argument of a complex number and how is it related to the real and imaginary parts?
3. How do you convert a complex number from rectangular to polar form?
4. Why is the argument \(315^\circ\) in this case and not \(45^\circ\)?
5. What are the uses of trigonometric form in complex number calculations?

**Tip:** Always check the quadrant of the complex number when determining the argument to ensure the angle is correct.

Write the complex number 5 - 5i in trigonometric form r( cos θ + i sin θ), with θ in the interval [0°,360°).

Learn how to convert the complex number 5 - 5i into its trigonometric form using the modulus and argument formulas. This solution involves finding the modulus, r = 5√2, and the argument, θ = 315°, ensuring the correct representation in polar form. Ideal for students in Grades 10-12.

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