Math Problem Statement

I^52

Solution

I52I^{52}, where II represents the imaginary unit, can be simplified using the fact that II has a cyclic property. The powers of II repeat every four terms:

I1=I,I2=1,I3=I,I4=1I^1 = I,\quad I^2 = -1,\quad I^3 = -I,\quad I^4 = 1

To simplify I52I^{52}, divide 52 by 4 and consider the remainder:

52÷4=13 (quotient) with remainder 052 \div 4 = 13 \text{ (quotient) with remainder } 0

Since the remainder is 0, we use the fact that I4=1I^4 = 1, so:

I52=1I^{52} = 1

Thus, I52=1I^{52} = 1.

Would you like more details or have any questions?

Here are 5 related questions to explore:

  1. What is the general form of powers of II?
  2. How can you apply this cyclic pattern to powers of other imaginary numbers?
  3. What are the applications of imaginary numbers in physics?
  4. How does Euler's formula connect to the powers of II?
  5. How would you simplify I100I^{100}?

Tip: When simplifying powers of imaginary numbers, always look for patterns in smaller powers.

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Math Problem Analysis

Mathematical Concepts

Imaginary Numbers
Cyclic Patterns
Exponentiation

Formulas

I^1 = I
I^2 = -1
I^3 = -I
I^4 = 1
I^n = I^(n mod 4)

Theorems

Cyclic properties of imaginary unit powers

Suitable Grade Level

Grades 9-12