Finding a Complex Conjugate
A polynomial function has [tex]$^{-5+\sqrt{3 i}}$[/tex] as a root. Which of the following must also be?
A. [tex]$^{-5-\sqrt{3 i}}$[/tex]
B. [tex]$^{-5+\sqrt{3 i}}$[/tex]
C. [tex]$^{5-\sqrt{3 i}}$[/tex]
D. [tex]$^{5+\sqrt{3 i}}$[/tex]
Answer (1)
The problem states that − 5 + 3 i is a root of a polynomial function.
We determine that if the polynomial has real coefficients, the conjugate of the root must also be a root.
We find the square root of 3 i to be 2 3 + i 2 3 .
Since the problem does not specify real coefficients, the only root that must also be a root is the given root itself: − 5 + 3 i .
Explanation
Understanding the Problem Let's analyze the given information. We have a polynomial function with a root of − 5 + 3 i . We need to determine which of the given options must also be a root. The key here is understanding complex conjugates and when they apply to polynomial roots.
Finding the Square Root of 3i If the polynomial has real coefficients, then complex roots occur in conjugate pairs. However, the given root involves the square root of a complex number, which complicates things. Let's first find out what 3 i actually is.
Calculating the Real and Imaginary Parts Let z = 3 i . Then z 2 = 3 i . We can express z as a + bi , where a and b are real numbers. Thus, ( a + bi ) 2 = a 2 + 2 abi − b 2 = 3 i . Equating the real and imaginary parts, we get a 2 − b 2 = 0 and 2 ab = 3 . From the first equation, a 2 = b 2 , so a = b or a = − b . If a = b , then 2 a 2 = 3 , so a = b = 2 3 . If a = − b , then − 2 a 2 = 3 , which is impossible since a is real. Therefore, 3 i = 2 3 + i 2 3 .
Finding the Complex Conjugate Now we know that the given root is − 5 + 2 3 + i 2 3 . If the polynomial has real coefficients, then its complex conjugate must also be a root. The complex conjugate is − 5 + 2 3 − i 2 3 . This can also be written as − 5 + 3 i .
Calculating the Conjugate of the Square Root Let's find 3 i . Since ( 3 i ) 2 = 3 i , we have ( 3 i ) 2 = 3 i = − 3 i . Let w = 3 i . Then w 2 = − 3 i . Let w = a + bi . Then ( a + bi ) 2 = a 2 − b 2 + 2 abi = − 3 i . So a 2 − b 2 = 0 and 2 ab = − 3 . Thus a = b or a = − b . If a = b , then 2 a 2 = − 3 , which is impossible. If a = − b , then − 2 a 2 = − 3 , so a 2 = 2 3 , and a = 2 3 and b = − 2 3 . Thus, 3 i = 2 3 − i 2 3 .
Considering Non-Real Coefficients The complex conjugate of the given root is − 5 + 2 3 − i 2 3 . This is equal to − 5 + 3 i , which is also equal to − 5 − − 3 i . Therefore, if the polynomial has real coefficients, then − 5 − − 3 i must also be a root. However, the problem does not state that the polynomial has real coefficients. Therefore, we cannot assume that the conjugate is also a root.
Final Answer Since we are given that − 5 + 3 i is a root, and we are asked which of the following must also be a root, the only option that must be a root is − 5 + 3 i itself. This is because if we don't have any information about the coefficients, we can't assume anything about other roots.
Conclusion Therefore, the answer is − 5 + 3 i .
Examples
Understanding complex roots is crucial in various fields like electrical engineering, where alternating current circuits are analyzed using complex numbers. For instance, when designing a circuit, engineers use complex impedance to represent the resistance and reactance. If the impedance equation has a complex root like − 5 + 3 i , knowing that its conjugate might also be a root helps in predicting the circuit's behavior and stability. This ensures the circuit functions as intended and avoids unwanted oscillations or failures. The concept of complex conjugates helps in simplifying calculations and understanding the symmetry in such systems.
