Math Problem Statement

how to simplify imaginary numbers

Solution

Imaginary numbers are simplified by applying the basic properties of i, where ii is defined as the square root of -1 (i=1i = \sqrt{-1}).

Here’s a breakdown of the key concepts and steps for simplifying expressions involving imaginary numbers:

1. Understanding Powers of ii:

The powers of ii follow a repeating cycle:

  • i1=ii^1 = i
  • i2=1i^2 = -1
  • i3=ii^3 = -i
  • i4=1i^4 = 1
  • This pattern repeats for higher powers of ii. So, i5=ii^5 = i, i6=1i^6 = -1, and so on.

2. Simplifying Expressions Involving Powers of ii:

For any power of ii, reduce the exponent modulo 4. For example:

  • i19i^{19} can be simplified by dividing 19 by 4. The remainder is 3, so i19=i3=ii^{19} = i^3 = -i.
  • i100i^{100} can be simplified by dividing 100 by 4. The remainder is 0, so i100=i0=1i^{100} = i^0 = 1.

3. Simplifying Expressions Involving Complex Numbers:

A complex number is of the form a+bia + bi, where aa and bb are real numbers.

  • To simplify an expression like (2+3i)+(45i)(2 + 3i) + (4 - 5i), you simply add the real parts and the imaginary parts separately: (2+3i)+(45i)=(2+4)+(3i5i)=62i(2 + 3i) + (4 - 5i) = (2 + 4) + (3i - 5i) = 6 - 2i
  • To simplify (2+3i)×(45i)(2 + 3i) \times (4 - 5i), use the distributive property (FOIL method): (2+3i)(45i)=2(4)+2(5i)+3i(4)+3i(5i)(2 + 3i)(4 - 5i) = 2(4) + 2(-5i) + 3i(4) + 3i(-5i) =810i+12i15i2= 8 - 10i + 12i - 15i^2 Since i2=1i^2 = -1, the expression becomes: 810i+12i+15=23+2i8 - 10i + 12i + 15 = 23 + 2i

4. Simplifying Expressions with ii in Denominators:

When you have an imaginary number in the denominator, multiply both the numerator and denominator by the conjugate of the denominator to simplify:

  • Example: Simplify 13+4i\frac{1}{3 + 4i}. Multiply by the conjugate of 3+4i3 + 4i, which is 34i3 - 4i: 13+4i×34i34i=34i(3+4i)(34i)\frac{1}{3 + 4i} \times \frac{3 - 4i}{3 - 4i} = \frac{3 - 4i}{(3 + 4i)(3 - 4i)} Use the formula (a+bi)(abi)=a2+b2(a + bi)(a - bi) = a^2 + b^2: (3+4i)(34i)=32+42=9+16=25(3 + 4i)(3 - 4i) = 3^2 + 4^2 = 9 + 16 = 25 So, the expression becomes: 34i25=3254i25\frac{3 - 4i}{25} = \frac{3}{25} - \frac{4i}{25}

5. Dealing with Higher Powers of ii:

If you encounter higher powers, always reduce the exponents as described earlier (modulo 4). This makes simplification easier.


Would you like to see an example for a more specific problem? Or do you have any other questions on simplifying imaginary numbers? Here's a list of related questions for further learning:

  1. How do you multiply two complex numbers?
  2. What is the conjugate of a complex number, and how does it simplify expressions?
  3. Can imaginary numbers be added to real numbers directly?
  4. What is the geometric interpretation of imaginary numbers in the complex plane?
  5. How do you handle complex numbers in polar form?

Tip: Always remember that powers of ii repeat in a cycle every 4 terms. This can make simplifying larger powers very efficient.

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Math Problem Analysis

Mathematical Concepts

Imaginary Numbers
Complex Numbers
Powers of i
Algebra

Formulas

i^2 = -1
(a + bi)(a - bi) = a^2 + b^2
i^4 = 1

Theorems

Properties of i
Simplification of Complex Numbers
Multiplication of Complex Numbers

Suitable Grade Level

Grades 9-12