Math Problem Statement

-i + (19 + 22i) - (8 - 14i) and -9i^65

Solution

Let's work through each question from the image.

Question 1:

Problem: Simplify and put your answer in standard form with NO SPACES. i+(19+22i)(814i)-i + (19 + 22i) - (8 - 14i)

Solution:

  1. Distribute the negative sign for the last term: i+(19+22i)8+14i-i + (19 + 22i) - 8 + 14i
  2. Now, combine the real parts and imaginary parts separately:
    • Real parts: 198=1119 - 8 = 11
    • Imaginary parts: i+22i+14i=35i-i + 22i + 14i = 35i
  3. Combine these results to get: 11+35i11 + 35i

Answer: 11+35i11+35i


Question 2:

Problem: Simplify 9i65-9i^{65}

Solution:

  1. Recall that the powers of ii repeat in a cycle 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
  2. To find i65i^{65}, we divide 65 by 4 and look at the remainder: 65÷4=16 remainder 165 \div 4 = 16 \text{ remainder } 1 So, i65=i1=ii^{65} = i^1 = i.
  3. Now, substitute back: 9i65=9i-9i^{65} = -9i

Answer: 9i-9i


Do you want more details on these solutions or have any questions?

Here are some related questions for further practice:

  1. How do you simplify expressions with complex numbers?
  2. What is the pattern of powers of ii and why do they repeat every four terms?
  3. How do you add and subtract complex numbers?
  4. What does it mean for a complex number to be in standard form?
  5. How do you multiply complex numbers and simplify the result?

Tip: Remember that the powers of ii cycle every four terms, so you can reduce high powers by finding the remainder when divided by 4.

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Math Problem Analysis

Mathematical Concepts

Complex Numbers
Algebra
Powers of i

Formulas

Standard form of a complex number: a + bi
Cycle of powers of i: i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1

Theorems

Properties of Complex Numbers
Cyclic nature of powers of i

Suitable Grade Level

Grades 10-12