Write The Expression In The Standard Form $a + Bi$.$\left(\frac{1}{6}+\frac{\sqrt{19}}{6} I\right)^2$

Introduction

In mathematics, complex numbers are a fundamental concept that extends the real number system to include numbers with both real and imaginary parts. The standard form of a complex number is given by $a + bi$, where $a$ and $b$ are real numbers and $i$ is the imaginary unit, which satisfies $i^2 = -1$. In this article, we will explore how to square a complex number in standard form, specifically the expression $\left(\frac{1}{6}+\frac{\sqrt{19}}{6} i\right)^2$.

The Binomial Theorem

To square the complex number $\left(\frac{1}{6}+\frac{\sqrt{19}}{6} i\right)$, we can use the binomial theorem, which states that for any non-negative integer $n$,

$$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$

where $\binom{n}{k}$ is the binomial coefficient, defined as

$$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$

In our case, we have $a = \frac{1}{6}$, $b = \frac{\sqrt{19}}{6} i$, and $n = 2$. Plugging these values into the binomial theorem, we get

$$\left(\frac{1}{6}+\frac{\sqrt{19}}{6} i\right)^2 = \binom{2}{0} \left(\frac{1}{6}\right)^2 + \binom{2}{1} \left(\frac{1}{6}\right) \left(\frac{\sqrt{19}}{6} i\right) + \binom{2}{2} \left(\frac{\sqrt{19}}{6} i\right)^2$$

Expanding the Binomial

Now, let's expand the binomial using the binomial coefficients and the values of $a$ and $b$.

$$\left(\frac{1}{6}+\frac{\sqrt{19}}{6} i\right)^2 = \left(\frac{1}{6}\right)^2 + 2 \left(\frac{1}{6}\right) \left(\frac{\sqrt{19}}{6} i\right) + \left(\frac{\sqrt{19}}{6} i\right)^2$$

Simplifying each term, we get

$$\left(\frac{1}{6}+\frac{\sqrt{19}}{6} i\right)^2 = \frac{1}{36} + \frac{2\sqrt{19}}{36} i + \frac{19}{36} i^2$$

Simplifying the Expression

Now, let's simplify the expression by combining like terms and using the fact that $i^2 = -1$.

$$\left(\frac{1}{6}+\frac{\sqrt{19}}{6} i\right)^2 = \frac{1}{36} + \frac{2\sqrt{19}}{36} i - \frac{19}{36}$$

Combining the real and imaginary parts, we get

$$\left(\frac{1}{6}+\frac{\sqrt{19}}{6} i\right)^2 = -\frac{18}{36} + \frac{2\sqrt{19}}{36} i$$

Simplifying further, we get

$$\left(\frac{1}{6}+\frac{\sqrt{19}}{6} i\right)^2 = -\frac{1}{2} + \frac{\sqrt{19}}{18} i$$

Conclusion

In this article, we have shown how to square a complex number in standard form using the binomial theorem. We have also simplified the resulting expression by combining like terms and using the fact that $i^2 = -1$. The final answer is $\boxed{-\frac{1}{2} + \frac{\sqrt{19}}{18} i}$.

Key Takeaways

• The binomial theorem can be used to square a complex number in standard form.
• The resulting expression can be simplified by combining like terms and using the fact that $i^2 = -1$.
• The final answer is $\boxed{-\frac{1}{2} + \frac{\sqrt{19}}{18} i}$.

Further Reading

If you want to learn more about complex numbers and the binomial theorem, I recommend checking out the following resources:

• Wikipedia: Complex Numbers
• Wikipedia: Binomial Theorem
• Khan Academy: Complex Numbers
• Khan Academy: Binomial Theorem
  Q&A: Squaring Complex Numbers in Standard Form =====================================================

Introduction

In our previous article, we explored how to square a complex number in standard form using the binomial theorem. In this article, we will answer some frequently asked questions about squaring complex numbers in standard form.

Q: What is the standard form of a complex number?

A: The standard form of a complex number is given by $a + bi$, where $a$ and $b$ are real numbers and $i$ is the imaginary unit, which satisfies $i^2 = -1$.

Q: How do I square a complex number in standard form?

A: To square a complex number in standard form, you can use the binomial theorem, which states that for any non-negative integer $n$,

$$(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$$

Q: What is the binomial theorem?

A: The binomial theorem is a mathematical formula that describes the expansion of a binomial raised to a power. It is a powerful tool for expanding expressions of the form $(a + b)^n$.

Q: How do I simplify the resulting expression after squaring a complex number?

A: To simplify the resulting expression after squaring a complex number, you can combine like terms and use the fact that $i^2 = -1$. This will help you to simplify the expression and write it in the standard form $a + bi$.

Q: What is the final answer for the expression $\left(\frac{1}{6}+\frac{\sqrt{19}}{6} i\right)^2$?

A: The final answer for the expression $\left(\frac{1}{6}+\frac{\sqrt{19}}{6} i\right)^2$ is $\boxed{-\frac{1}{2} + \frac{\sqrt{19}}{18} i}$.

Q: Can I use the binomial theorem to square a complex number with a negative imaginary part?

A: Yes, you can use the binomial theorem to square a complex number with a negative imaginary part. The binomial theorem works for any non-negative integer $n$, so you can use it to square a complex number with a negative imaginary part.

Q: Are there any other ways to square a complex number in standard form?

A: Yes, there are other ways to square a complex number in standard form. One way is to use the formula $(a + bi)^2 = a^2 + 2abi - b^2$, which is derived from the binomial theorem.

Conclusion

In this article, we have answered some frequently asked questions about squaring complex numbers in standard form. We have also provided some additional information and resources for further learning.

Key Takeaways

• The standard form of a complex number is given by $a + bi$.
• The binomial theorem can be used to square a complex number in standard form.
• The resulting expression can be simplified by combining like terms and using the fact that $i^2 = -1$.
• The final answer for the expression $\left(\frac{1}{6}+\frac{\sqrt{19}}{6} i\right)^2$ is $\boxed{-\frac{1}{2} + \frac{\sqrt{19}}{18} i}$.

Further Reading

If you want to learn more about complex numbers and the binomial theorem, I recommend checking out the following resources:

• Wikipedia: Complex Numbers
• Wikipedia: Binomial Theorem
• Khan Academy: Complex Numbers
• Khan Academy: Binomial Theorem