assignment 6: problem 13
(1 point)
evaluate the expression $ \frac { - 1 + \sqrt { - 16 } } { - 1 + \sqrt { - 1 } } $ and write the result in the form $ a + b i $.
the real number $ a $ equals
the real number $ b $ equals
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Accepted Answer

# Explanation:
## Step1: Simplify the square roots
We know that $ \sqrt{-1}=i $ and $ \sqrt{-16} = 4i $.
So the expression $ \frac{-1+\sqrt{-16}}{-1+\sqrt{-1}}=\frac{-1 + 4i}{-1+i} $.
## Step2: Rationalize the denominator
Multiply the numerator and denominator by the conjugate of the denominator. The conjugate of $ -1 + i $ is $ -1 - i $.
$$
\begin{align*}
\frac{(-1 + 4i)(-1 - i)}{(-1 + i)(-1 - i)}&=\frac{(-1)(-1)+(-1)(-i)+4i(-1)+4i(-i)}{(-1)^2-(i)^2}\
&=\frac{1 + i-4i-4i^{2}}{1 - (-1)}\
\end{align*}
$$
Since $ i^{2}=-1 $, we have:
$$
\begin{align*}
\frac{1 + i-4i-4(-1)}{1 + 1}&=\frac{1 + i-4i + 4}{2}\
&=\frac{(1 + 4)+(i-4i)}{2}\
&=\frac{5-3i}{2}\
&=\frac{5}{2}-\frac{3}{2}i
\end{align*}
$$

# Answer:
The real number $ a=\frac{5}{2} $, the real number $ b =-\frac{3}{2} $

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QUESTION IMAGE

Question

Explanation:

Step1: Simplify the square roots

Step2: Rationalize the denominator

Answer: