Suppose a and b are real numbers, not both 0. Find real numbers c and d such that

1/(a +bi) = c+di

To find real numbers \( c \) and \( d \) such that:

\[
\frac{1}{a + bi} = c + di
\]

we can multiply both sides of the equation by the complex conjugate of the denominator to eliminate the imaginary part in the denominator.

### Step 1: Multiply by the Complex Conjugate

The complex conjugate of \( a + bi \) is \( a - bi \). Multiplying both the numerator and the denominator by this conjugate gives:

\[
\frac{1}{a + bi} \cdot \frac{a - bi}{a - bi} = \frac{a - bi}{(a + bi)(a - bi)}
\]

### Step 2: Simplify the Denominator

The denominator is a difference of squares:

\[
(a + bi)(a - bi) = a^2 - (bi)^2
\]

Since \( i^2 = -1 \), this becomes:

\[
a^2 - (-b^2) = a^2 + b^2
\]

So, the expression simplifies to:

\[
\frac{a - bi}{a^2 + b^2}
\]

### Step 3: Separate Real and Imaginary Parts

The expression \( \frac{a - bi}{a^2 + b^2} \) can be written as:

\[
\frac{a}{a^2 + b^2} - \frac{b}{a^2 + b^2}i
\]

Comparing this with \( c + di \), we have:

\[
c = \frac{a}{a^2 + b^2}, \quad d = -\frac{b}{a^2 + b^2}
\]

### Final Answer

The real numbers \( c \) and \( d \) that satisfy the equation \( \frac{1}{a + bi} = c + di \) are:

\[
c = \frac{a}{a^2 + b^2}, \quad d = -\frac{b}{a^2 + b^2}
\]

Would you like more details on this, or do you have any questions?

Here are some related questions you might find interesting:

1. How does the formula for \( c \) and \( d \) change if \( a \) or \( b \) is zero?
2. Can you derive the same result by using the polar form of a complex number?
3. How would you interpret the result geometrically in the complex plane?
4. How would the formula change if we considered a different complex number instead of \( 1 \)?
5. What is the significance of the modulus \( \sqrt{a^2 + b^2} \) in the context of this problem?

**Tip:** Multiplying by the complex conjugate is a powerful technique for simplifying complex fractions and is essential when dividing complex numbers.

Learn how to find real numbers c and d such that 1/(a + bi) = c + di. Step-by-step solution involving complex conjugates and fraction simplification. Suitable for advanced high school students.

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