How do I find (left(frac{1-i}{1i}right)^{100})?
Understanding complex numbers and how to manipulate them using polar coordinates and De Moivre's theorem is a crucial skill in many fields of mathematics and engineering. In this article, we will explore the process of finding (left(frac{1-i}{1i}right)^{100}) and provide a step-by-step guide to solving this problem.
Step 1: Simplify (frac{1-i}{1i}) in Polar Form
First, let's convert the complex number (1-i) to its polar form. The polar form of a complex number (a bi) is given by (r(costheta isintheta)), where (r) is the modulus and (theta) is the argument.
The Modulus and Argument of (1-i)
The modulus (r) is calculated as follows:
[r |1-i| sqrt{1^2 (-1)^2} sqrt{2}]The argument (theta) is given by:
[theta_1 tan^{-1}left(frac{-1}{1}right) -frac{pi}{4}]Thus, the polar form of (1-i) is:
[1-i sqrt{2}(cos(-frac{pi}{4}) isin(-frac{pi}{4}))]The Modulus and Argument of (1i)
For the complex number (1i), the modulus is:
[r |1i| sqrt{1^2 1^2} sqrt{2}]The argument (theta) is given by:
[theta_2 tan^{-1}left(frac{1}{1}right) frac{pi}{4}]Thus, the polar form of (1i) is:
[1i sqrt{2}(cos(frac{pi}{4}) isin(frac{pi}{4}))]Quotient of Two Complex Numbers
To find (frac{1-i}{1i}), we divide the polar forms of the two complex numbers:
[frac{1-i}{1i} frac{sqrt{2}(cos(-frac{pi}{4}) isin(-frac{pi}{4}))}{sqrt{2}(cos(frac{pi}{4}) isin(frac{pi}{4}))} cos(-frac{pi}{4} - frac{pi}{4}) isin(-frac{pi}{4} - frac{pi}{4})]Simplifying the arguments:
[cos(-frac{pi}{2}) isin(-frac{pi}{2}) -i]Step 2: Raise to the Power of 100 Using De Moivre's Theorem
Now that we have (frac{1-i}{1i} -i), we can raise it to the power of 100:
[left(-iright)^{100} left(e^{-ifrac{pi}{2}}right)^{100} e^{-i50pi}]Simplifying (e^{-i50pi})
Using the properties of the exponential function, we know:
[e^{-i50pi} cos(50pi) isin(50pi)]Since (50pi) is an even multiple of (pi), we have:
[cos(50pi) 1 quad text{and} quad sin(50pi) 0]Thus:
[e^{-i50pi} 1 0i 1]Final Result
The final result is:
[left(frac{1-i}{1i}right)^{100} 1]Alternative Methods
There are alternative methods to solving this problem. For example, we can use the fact that (left(frac{1-i}{1i}right)^2 -1). Therefore:
[left(frac{1-i}{1i}right)^{100} left(left(frac{1-i}{1i}right)^2right)^{50} (-1)^{50} 1]This confirms our result.
A Mathematical Insight
Let (y left(frac{1-i}{1i}right)^{100}). Let (x frac{1-i}{1i} frac{1-i}{1i} cdot frac{1-i}{1-i} -i). Therefore:
[y x^{100} -i^{100} i^4^{25} 1^{25} 1]Thus, the polar form of the result is:
[rtheta 1 0i]This confirms that (left(frac{1-i}{1i}right)^{100} 1).
In conclusion, understanding complex numbers and De Moivre's theorem provides a powerful tool for solving problems involving complex exponentiation. This method can be applied to a wide range of problems in advanced mathematics and engineering.