Respuesta :
Answer:
Step-by-step explanation:
To solve for \( k \), we'll rationalize the denominator of the fraction:
\[
\frac{k - 1 + \sqrt{3}i}{-5\sqrt{3} + 5i} = ki
\]
First, let's rationalize the denominator by multiplying both the numerator and denominator by the conjugate of the denominator:
\[
\frac{(k - 1 + \sqrt{3}i)(-5\sqrt{3} - 5i)}{(-5\sqrt{3} + 5i)(-5\sqrt{3} - 5i)} = ki
\]
This simplifies to:
\[
\frac{-5k\sqrt{3} - 5ki + 5\sqrt{3} - 5i - 5\sqrt{3}i - 5}{75 + 75} = ki
\]
\[
\frac{-5k\sqrt{3} - 5ki - 5i - 5}{150} = ki
\]
Now, let's separate the real and imaginary parts:
Real Part:
\[
\frac{-5k\sqrt{3} - 5}{150} = 0
\]
\[
-5k\sqrt{3} - 5 = 0
\]
\[
-5k\sqrt{3} = 5
\]
\[
k\sqrt{3} = -1
\]
\[
k = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}
\]
Imaginary Part:
\[
\frac{-5ki - 5i}{150} = k
\]
\[
-5ki - 5i = 150k
\]
\[
(-5i - 150)k = 5i
\]
\[
k = \frac{5i}{-5i - 150}
\]
\[
k = \frac{5i}{-5(i + 30)}
\]
\[
k = \frac{-1}{i + 30}
\]
Since \( k = \frac{\sqrt{3}}{3} \) and \( k = \frac{-1}{i + 30} \) both hold, they must be equal. Therefore:
\[
\frac{\sqrt{3}}{3} = \frac{-1}{i + 30}
\]
Now, multiply both sides by \( i + 30 \):
\[
\sqrt{3}(i + 30) = -3
\]
\[
\sqrt{3}i + 30\sqrt{3} = -3
\]
\[
\sqrt{3}i = -3 - 30\sqrt{3}
\]
\[
i = \frac{-3 - 30\sqrt{3}}{\sqrt{3}}
\]
\[
i = -\sqrt{3} - 30
\]
So, \( k = \frac{\sqrt{3}}{3} = -\sqrt{3} - 30 \).
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