Question

Find one or more values of each of the following complex expressions and compare with a computer solution. $$\left(\frac{1+i \sqrt{3}}{\sqrt{2}+i \sqrt{2}}\right)^{50}$$

Short answer

The final simplified result is obtained using De Moivre's theorem and represents the solution in polar form.

Step-by-step solution

- Simplify the Complex Fraction

Start by simplifying the complex fraction \(\frac{1+i \sqrt{3}}{\sqrt{2}+i \sqrt{2}}\). Multiply numerator and denominator by the conjugate of the denominator, which is \(\sqrt{2} - i \sqrt{2}\).

- Multiply and Simplify

Calculate the numerator and the denominator after multiplying with the conjugate: \[ (1+i \sqrt{3})(\sqrt{2}-i \sqrt{2}) \text{ and } (\sqrt{2}+i \sqrt{2})(\sqrt{2}-i \sqrt{2}) \] Simplify the expressions: \[ \text{Numerator: } 1(\sqrt{2}) + 1(-i \sqrt{2}) + i \sqrt{3} (\sqrt{2}) + i \sqrt{3} (-i \sqrt{2}) \text{and Denominator: } 2+2=4 \]

- Simplify Further

Combine like terms in the numerator: \[ \sqrt{2} - i \sqrt{2} + i \sqrt{6} + \sqrt{6} \] Simplify to get: \[ ( \sqrt{2} + \sqrt{6} + i( \sqrt{6} - \sqrt{2}) )/4 \]

- Convert to Polar Form

Convert \(\frac{\sqrt{2} + \sqrt{6} + i \sqrt{6} - i \sqrt{2}}{4}\) to polar form. Find \( r\) and \( \theta \): \ r = \sqrt{a^2+b^2}, \theta = \arctan \(b/a\)

- Raise to the Power 50

Use De Moivre's Theorem to raise the complex number to the power 50: \[ \left( r e^{i \theta} \right)^{50} = r^{50} e^{i \theta \cdot 50} \]

- Simplify the Result

Simplify the exponent and convert back to rectangular form for the final result.

Key concepts

Simplifying Complex Fractions

Simplifying complex fractions might seem intimidating, but with a systematic approach, it becomes manageable. The key is to eliminate the complex denominator by multiplying both the numerator and the denominator by the conjugate of the denominator. The conjugate of a complex number is found by changing the sign of the imaginary part. For example, the conjugate of \[ \sqrt{2} + i \sqrt{2} \] is \[ \sqrt{2} - i \sqrt{2} \]. By using the conjugate, we convert the denominator into a real number. After multiplying, we'll need to simplify both the numerator and the denominator further to get a final simplified form.

Conjugates in Complex Numbers

Conjugates play a crucial role when working with complex numbers, particularly in simplifying fractions and performing division. The conjugate of a complex number \[a + bi\] is \[a - bi\]. When you multiply a complex number by its conjugate, you get a real number: \[(a + bi)(a - bi) = a^2 + b^2\], effectively eliminating the imaginary part. This property is useful in simplifying complex fractions, as demonstrated in the solution above. Multiplying by the conjugate simplifies the problem by converting the denominator into a real number, making it easier to handle.

De Moivre's Theorem

De Moivre's Theorem is an essential tool for raising complex numbers to a power. It states that for a complex number in polar form, \[ \left( r e^{i \theta} \right)^n = r^n e^{i n \theta} \], where \[n\] is any real number. This theorem simplifies the process significantly. To apply it, first convert your complex number into polar form, which involves finding the magnitude \[r\] and the argument \[ \theta \]. Then, simply raise the magnitude to the desired power and multiply the argument by the power. Finally, convert back to rectangular form if needed.

Polar Form of Complex Numbers

The polar form of a complex number provides another way to represent complex numbers, particularly useful in multiplication, division, and exponentiation. A complex number \[ z = a + bi \] can be converted to polar form as \[ z = r (\cos \theta + i \sin \theta ) \], where \[ r = \sqrt{a^2 + b^2} \] is the magnitude and \[ \theta = \arctan \frac{b}{a} \] is the argument. Once in polar form, operations such as raising a complex number to a power or multiplying/dividing become much simpler, thanks to De Moivre’s Theorem and other properties of exponents.