Verified Answer :

  • Identify the cyclic powers of \(i\) where \(i^1=i\), \(i^2=-1\), \(i^3=-i\), and \(i^4=1\).
  • Simplify each term by calculating the exponent modulo 4: \(i^{51}=-i\), \(i^{115}=-i\), \(i^{76}=1\), and \(i^{114}=-1\).
  • Substitute these into the expression: \(2(-i) + 10(-i) + 7(1) + 9(-1)\).
  • Combine the resulting real and imaginary parts to obtain \(\boxed{-2 - 12i}\).

Explanation

  1. Analysis of the Expression and Imaginary Unit Powers

The expression to simplify is \(2i^{51} + 10i^{115} + 7i^{76} + 9i^{114}\). The imaginary unit \(i\) is defined by the property \(i^2 = -1\). Its powers follow a cyclic pattern of period 4:

  • \(i^1 = i\)
  • \(i^2 = -1\)
  • \(i^3 = -i\)
  • \(i^4 = 1\)

Because \(i^4 = 1\), any power \(i^n\) can be simplified by evaluating \(n \pmod 4\). Specifically, \(i^n = i^{n \pmod 4}\), where \(i^0\) is interpreted as \(1\). We will use this to simplify each term individually.

  1. Calculating Modular Exponents

We determine the remainder of each exponent when divided by 4:

  • For \(i^{51}\): \(51 = 4 \times 12 + 3\), so \(51 \equiv 3 \pmod 4\). Thus, \(i^{51} = i^3 = -i\).
  • For \(i^{115}\): \(115 = 4 \times 28 + 3\), so \(115 \equiv 3 \pmod 4\). Thus, \(i^{115} = i^3 = -i\).
  • For \(i^{76}\): \(76 = 4 \times 19 + 0\), so \(76 \equiv 0 \pmod 4\). Thus, \(i^{76} = i^0 = 1\).
  • For \(i^{114}\): \(114 = 4 \times 28 + 2\), so \(114 \equiv 2 \pmod 4\). Thus, \(i^{114} = i^2 = -1\).
  1. Simplifying and Combining Terms

Substituting these simplified forms back into the original expression:
\(2i^{51} + 10i^{115} + 7i^{76} + 9i^{114} = 2(-i) + 10(-i) + 7(1) + 9(-1)\).
Now perform the arithmetic:

  • \(= -2i - 10i + 7 - 9\)
  • Combine the imaginary parts: \(-2i - 10i = -12i\)
  • Combine the real parts: \(7 - 9 = -2\)

Thus, the expression simplifies to \(-2 - 12i\).

Examples

In electrical engineering, complex numbers are fundamental for analyzing alternating current (AC) circuits. When dealing with resistors, capacitors, and inductors in a circuit, we represent their impedance as complex numbers. Adding these impedances requires performing arithmetic with complex numbers, such as simplifying expressions like the one above, to determine the total resistance or "complex impedance" of the system, which directly affects the current flow.