Math Problem Statement

(1+2i)(2-3i)/((2-i)(3+2i))

Solution

To solve the given expression:

(1+2i)(23i)(2i)(3+2i)\frac{(1+2i)(2-3i)}{(2-i)(3+2i)}

we'll simplify the numerator and denominator separately, then simplify the overall expression.

Step 1: Simplifying the numerator (1+2i)(23i)(1+2i)(2-3i)

We can expand this by using the distributive property (FOIL):

(1+2i)(23i)=1(2)+1(3i)+2i(2)+2i(3i)(1+2i)(2-3i) = 1(2) + 1(-3i) + 2i(2) + 2i(-3i)

Simplify:

=23i+4i6i2= 2 - 3i + 4i - 6i^2

Recall that i2=1i^2 = -1, so:

=23i+4i+6=8+i= 2 - 3i + 4i + 6 = 8 + i

Thus, the numerator simplifies to:

8+i8 + i

Step 2: Simplifying the denominator (2i)(3+2i)(2-i)(3+2i)

Again, we expand using the distributive property:

(2i)(3+2i)=2(3)+2(2i)i(3)i(2i)(2-i)(3+2i) = 2(3) + 2(2i) - i(3) - i(2i)

Simplify:

=6+4i3i2i2= 6 + 4i - 3i - 2i^2

Using i2=1i^2 = -1:

=6+4i3i+2=8+i= 6 + 4i - 3i + 2 = 8 + i

Thus, the denominator simplifies to:

8+i8 + i

Step 3: Putting it together

Now we can rewrite the entire expression:

(1+2i)(23i)(2i)(3+2i)=8+i8+i\frac{(1+2i)(2-3i)}{(2-i)(3+2i)} = \frac{8+i}{8+i}

Since both the numerator and denominator are the same, the expression simplifies to:

11

Final Answer:

1\boxed{1}


Would you like a detailed breakdown of any specific step? Or have any other questions related to complex numbers?

Here are 5 related questions to explore:

  1. How do you handle the multiplication of complex numbers using polar form?
  2. Can you explain how to divide complex numbers using conjugates?
  3. What happens if you multiply a complex number by its conjugate?
  4. What is the geometric interpretation of multiplying complex numbers?
  5. How can you represent complex numbers on the complex plane?

Tip: When multiplying or dividing complex numbers, always remember to use the distributive property carefully and simplify i2i^2 as 1-1.

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Math Problem Analysis

Mathematical Concepts

Complex Numbers
Multiplication of Complex Numbers
Division of Complex Numbers
Simplification of Expressions

Formulas

(a+bi)(c+di) = (ac-bd) + (ad+bc)i
(a+bi)/(c+di) = ((a+bi)(c-di))/((c+di)(c-di))

Theorems

Multiplying Complex Numbers
Division of Complex Numbers using Conjugates

Suitable Grade Level

Grades 9-12