Math  /  Algebra

QuestionSimplify. 89i\frac{-8}{9-i}
Write your answer in the form a + bi. Reduce all fractions. \square

Studdy Solution

STEP 1

What is this asking? We need to simplify a fraction with a complex number in the denominator and write the answer in the standard *a* + *bi* form. Watch out! Remember that *i* is the square root of -1, and *i* squared is -1.
Don't forget to multiply both the numerator and denominator by the conjugate!

STEP 2

1. Find the conjugate
2. Multiply by the conjugate
3. Simplify

STEP 3

The **conjugate** of a complex number *a* + *bi* is *a* - *bi*.
Basically, we just flip the sign of the imaginary part.

STEP 4

In our problem, the denominator is 9i9 - i.
So, the **conjugate** is 9+i9 + i.
It's that simple!

STEP 5

To get rid of the complex number in the denominator, we'll **multiply** both the numerator and the denominator of our original fraction by the conjugate we just found.
This is like multiplying by 1, so it doesn't change the value of the fraction, just how it looks!

STEP 6

Here's how we do it: 89i9+i9+i \frac{-8}{9-i} \cdot \frac{9+i}{9+i}

STEP 7

Let's **multiply** the numerators: 8(9+i)=728i -8 \cdot (9 + i) = -72 - 8i

STEP 8

Now, let's **multiply** the denominators: (9i)(9+i)=81+9i9ii2 (9 - i) \cdot (9 + i) = 81 + 9i - 9i - i^2 Since i2=1i^2 = -1, we have: 81+9i9i(1)=81+1=82 81 + 9i - 9i - (-1) = 81 + 1 = 82 So, our fraction now looks like this: 728i82 \frac{-72 - 8i}{82}

STEP 9

We can **separate** this fraction into two parts: 7282882i \frac{-72}{82} - \frac{8}{82}i

STEP 10

Now, we **reduce** each fraction.
Both the numerator and denominator of the first fraction are divisible by 2, and the same is true for the second fraction. 7282=72÷282÷2=3641 \frac{-72}{82} = \frac{-72 \div 2}{82 \div 2} = \frac{-36}{41} 882=8÷282÷2=441 \frac{8}{82} = \frac{8 \div 2}{82 \div 2} = \frac{4}{41}

STEP 11

Putting it all together, our **simplified** complex number is: 3641441i -\frac{36}{41} - \frac{4}{41}i

STEP 12

Our final answer is 3641441i-\frac{36}{41} - \frac{4}{41}i.

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