Math Problem Statement

Express 2 (2-i)(2+I)-1 in the form of a+ib

Solution

We need to simplify the expression 2(2i)(2+i)12(2 - i)(2 + i) - 1 and express the result in the form a+iba + ib, where aa and bb are real numbers.

Step 1: Simplify (2i)(2+i)(2 - i)(2 + i)

Notice that (2i)(2+i)(2 - i)(2 + i) is a difference of squares. We can apply the formula: (ab)(a+b)=a2b2(a - b)(a + b) = a^2 - b^2 Here, a=2a = 2 and b=ib = i, so: (2i)(2+i)=22i2=4(1)=4+1=5(2 - i)(2 + i) = 2^2 - i^2 = 4 - (-1) = 4 + 1 = 5

Step 2: Multiply by 2

Now we multiply the result by 2: 2(2i)(2+i)=2×5=102(2 - i)(2 + i) = 2 \times 5 = 10

Step 3: Subtract 1

Finally, subtract 1 from the result: 2(2i)(2+i)1=101=92(2 - i)(2 + i) - 1 = 10 - 1 = 9

Final Answer:

The expression simplifies to 9+0i9 + 0i, which means the real part is 9 and the imaginary part is 0. Therefore, the expression in the form a+iba + ib is: 9+0i9 + 0i

Would you like more details or have any questions?

Here are five related questions to deepen your understanding:

  1. What is the difference of squares, and why does it apply here?
  2. How does the value of i2i^2 influence calculations involving complex numbers?
  3. Can you verify that (2i)(2+i)=5(2 - i)(2 + i) = 5 by expanding directly?
  4. How would this calculation change if the imaginary unit ii was replaced by another constant?
  5. What are some other common operations involving complex numbers?

Tip: Remember, multiplying a complex number by its conjugate always results in a real number!

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Math Problem Analysis

Mathematical Concepts

Complex Numbers
Algebra
Difference of Squares

Formulas

(a - b)(a + b) = a^2 - b^2

Theorems

Difference of Squares

Suitable Grade Level

Grades 9-12