Question

Evaluate each of the following in \(x+i y\) form, and compare with a computer solution. \((1-\sqrt{2 i})^{i} .\) Hint: Find \(\sqrt{2 i}\) first.

Short answer

The value is \(\boxed{e^{\frac{\pi}{2}}}\).

Step-by-step solution

Represent the Complex Number in Polar Form

First, transform the complex number \(2i\) into its polar form. A complex number can be expressed as \(re^{i\theta}\), where \(r\) is the magnitude and \(\theta\) is the argument. For the complex number \(2i\), the magnitude is \(2\) and the argument is \(\frac{\pi}{2}\). Therefore: \(2i = 2e^{i \frac{\pi}{2}}\).

Calculate the Square Root in Polar Form

Using the property of exponential functions, calculate the square root of \(2i\): \(sqrt{2i} = (2e^{i \frac{\pi}{2}})^{1/2} = \sqrt{2} e^{i \frac{\pi}{4}}\). Therefore, \(\sqrt{2i} = \sqrt{2} (\cos(\frac{\pi}{4}) + i \sin(\frac{\pi}{4}))\), which simplifies to: \(\sqrt{2i} = \sqrt{2}(\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}) = 1 + i\).

Represent \(1 - \sqrt{2i}\) in Terms of Known Quantities

Now substitute the value found for \(\sqrt{2i}\) into \(1 - \sqrt{2i}\). \(1 - \sqrt{2i} = 1 - (1 + i) = -i\).

Raise \(-i\) to the Power of \(i\)

Express \(-i\) in polar form: \(i\) has a magnitude of 1 and an argument of \(-\frac{\pi}{2}\) (since it lies on the negative imaginary axis). Therefore, \(-i = e^{-i \frac{\pi}{2}}\). Raising this to the power of \(i\): \((-i)^i = \left(e^{-i \frac{\pi}{2}}\right)^i = e^{-i^2 \frac{\pi}{2}} = e^{\frac{\pi}{2}}\).

Key concepts

polar form of complex numbers

The polar form is a way to express complex numbers in terms of their magnitude and direction.
It's useful particularly for multiplication and division operations. In polar form, a complex number is represented as:
\[ z = re^{i\theta} \]
Where:

• \r is the magnitude
• \theta is the argument (angle)

Take the complex number \(2i\), for example. Its magnitude is 2 and it lies on the positive imaginary axis. The angle it makes with the positive real axis is \(\frac{\pi}{2}\). So, in polar form:
\[ 2i = 2e^{i\frac{\pi}{2}} \]
This transformation simplifies many operations, such as finding roots or powers of complex numbers.

exponential function in complex numbers

The exponential function \(e\) plays a crucial role in the world of complex numbers. When dealing with complex numbers, the exponential form \(e^{i\theta}\) helps to simplify many operations.
The exponential form is connected to the Euler's formula:
\[ e^{i\theta} = \cos(\theta) + i\sin(\theta) \]
An example would be taking the square root of a complex number like \(2i\):

• Express \(2i\) in polar form as \(2e^{i\frac{\pi}{2}}\)
• Find the square root:
  \[\sqrt{2 e^{i\frac{\pi}{2}}} = \sqrt{2} e^{i\frac{\pi}{4}} \]

Using Euler's formula, you can split this back into cartesian form:
\[ \sqrt{2i} = \sqrt{2} \left(\cos\left(\frac{\pi}{4}\right) + i\sin\left(\frac{\pi}{4}\right)\right) = 1 + i \]

raising complex numbers to a power

Raising complex numbers to a power can be more straightforward when using their polar forms.
Once in polar form, exponentiation rules can be applied naturally. Consider the following steps:

• First, express the complex number in polar form. For instance, \(-i\) can be written as:
  \(-i = e^{-i\frac{\pi}{2}}\)
• Raise the complex number to a power. Here we're raising \(-i\) to the power of \(i\):\[(e^{-i\frac{\pi}{2}})^i \]
  Using properties of exponents:\
  \( e^{-i^2\frac{\pi}{2}} \)
  Since \(i^2 = -1\):\
  \( e^{\frac{\pi}{2}} \)
• This simplifies to a real number value:
  \( e^{\frac{\pi}{2}} \)

In this process, moving from cartesian to polar form makes exponentiation much simpler.