https://file.aiquickdraw.com/mathbot/file/172740480917350dvxcn5.jpg

The expression in the image is:

\[
\frac{4}{5 - 4i}
\]

To write this in the standard form \( a + bi \), we need to eliminate the imaginary number in the denominator. This can be done by multiplying both the numerator and denominator by the conjugate of the denominator, which in this case is \( 5 + 4i \).

Let's go through the steps:

### Step 1: Multiply by the conjugate
Multiply the numerator and denominator by \( 5 + 4i \):

\[
\frac{4}{5 - 4i} \times \frac{5 + 4i}{5 + 4i} = \frac{4(5 + 4i)}{(5 - 4i)(5 + 4i)}
\]

### Step 2: Simplify the denominator
The denominator becomes a difference of squares:

\[
(5 - 4i)(5 + 4i) = 5^2 - (4i)^2 = 25 - 16i^2 = 25 - 16(-1) = 25 + 16 = 41
\]

So, the denominator simplifies to 41.

### Step 3: Expand the numerator
Now, expand the numerator:

\[
4(5 + 4i) = 20 + 16i
\]

### Step 4: Combine the results
Now, put the simplified numerator over the denominator:

\[
\frac{20 + 16i}{41} = \frac{20}{41} + \frac{16i}{41}
\]

### Step 5: Final result
Thus, the expression in the form \( a + bi \) is:

\[
\frac{20}{41} + \frac{16}{41}i
\]

This is the simplified form.

Would you like further details or have any questions? Here are some related questions to deepen understanding:

1. How do you find the conjugate of a complex number?
2. What is the difference between real and imaginary parts in a complex number?
3. Why does multiplying by the conjugate help remove the imaginary part in the denominator?
4. Can you explain the difference of squares formula used in the denominator?
5. How would this problem change if the denominator were \( 5 + 4i \) instead of \( 5 - 4i \)?

**Tip:** When multiplying complex numbers, always remember that \( i^2 = -1 \), which helps simplify the expressions.

Write the expression \( \frac{4}{5 - 4i} \) in the standard form a + bi.

Learn how to write the complex expression \( \frac{4}{5 - 4i} \) in standard form a + bi by multiplying by the conjugate. Explore key concepts such as complex numbers, conjugates, and the difference of squares. This solution is suitable for high school students in Grades 10-12.

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