solve x^4+16=0 in complex number s using de moivren theory

To solve the equation \( x^4 + 16 = 0 \) using complex numbers and De Moivre's Theorem, follow these steps:

### Step 1: Rearrange the Equation
We start with the equation:
\[
x^4 + 16 = 0
\]
Rearrange it to:
\[
x^4 = -16
\]

### Step 2: Express \(-16\) in Polar Form
The complex number \(-16\) can be written in polar form. In Cartesian form, \(-16 = -16 + 0i\), which has a modulus \( r = 16 \) and an argument \( \theta = \pi \) (since \(-16\) lies on the negative real axis).

Thus, we express \(-16\) in polar form as:
\[
-16 = 16 \text{cis} \pi
\]
where \( \text{cis} \theta \) is shorthand for \( \cos \theta + i \sin \theta \).

### Step 3: Apply De Moivre's Theorem
We now apply De Moivre's Theorem to solve for \( x \). The equation \( x^4 = 16 \text{cis} \pi \) has four roots, given by:
\[
x = \left( 16 \text{cis} \pi \right)^{1/4}
\]

Taking the fourth root of the modulus and dividing the argument by 4, we get:
\[
x = 2 \text{cis} \left( \frac{\pi + 2k\pi}{4} \right), \quad k = 0, 1, 2, 3
\]
This accounts for the four possible solutions.

### Step 4: Calculate the Roots
Substitute values of \( k = 0, 1, 2, 3 \) to find the four roots.

1. For \( k = 0 \):
   \[
   x_1 = 2 \text{cis} \left( \frac{\pi}{4} \right) = 2 \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) = 2 \left( \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right) = \sqrt{2} + i\sqrt{2}
   \]

2. For \( k = 1 \):
   \[
   x_2 = 2 \text{cis} \left( \frac{3\pi}{4} \right) = 2 \left( \cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4} \right) = 2 \left( -\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right) = -\sqrt{2} + i\sqrt{2}
   \]

3. For \( k = 2 \):
   \[
   x_3 = 2 \text{cis} \left( \frac{5\pi}{4} \right) = 2 \left( \cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4} \right) = 2 \left( -\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \right) = -\sqrt{2} - i\sqrt{2}
   \]

4. For \( k = 3 \):
   \[
   x_4 = 2 \text{cis} \left( \frac{7\pi}{4} \right) = 2 \left( \cos \frac{7\pi}{4} + i \sin \frac{7\pi}{4} \right) = 2 \left( \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \right) = \sqrt{2} - i\sqrt{2}
   \]

### Final Answer:
The four complex roots of the equation \( x^4 + 16 = 0 \) are:
\[
x_1 = \sqrt{2} + i\sqrt{2}, \quad x_2 = -\sqrt{2} + i\sqrt{2}, \quad x_3 = -\sqrt{2} - i\sqrt{2}, \quad x_4 = \sqrt{2} - i\sqrt{2}
\]

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Do you have any questions, or would you like more details?

Here are 5 related questions:
1. How does De Moivre's Theorem work for solving general polynomial equations?
2. What is the significance of the argument in polar form for complex numbers?
3. Can you explain how the modulus of a complex number affects its roots?
4. How do we find the nth roots of a complex number?
5. What happens when a complex number has a modulus of 1 in polar form?

**Tip:** Always convert complex numbers to polar form when using De Moivre’s Theorem for finding roots—it simplifies handling both the modulus and the argument.

solve x^4+16=0 in complex numbers using de Moivre's theorem

Discover how to solve the equation x^4 + 16 = 0 using complex numbers and De Moivre's Theorem. Learn the step-by-step method to express -16 in polar form, apply De Moivre's theorem, and calculate the four complex roots. This guide is suitable for advanced high school students in Grades 11-12.

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