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The operation " \( \hat{\prime \prime} \) " on the set of complex numbers is defined as \( z_{1} \otimes z_{2}=z_{1}+z_{2}+\left|z_{1} \cdot z_{2}\right| \) What is the equivalent of the operation \( (1-2 i) *(2+i) \)

Ask by Stewart Ayala. in Qatar
Feb 03,2025

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Answer

The result of the operation \( (1 - 2i) \otimes (2 + i) \) is \( 8 - i \).

Solution

To evaluate the operation \( (1 - 2i) \otimes (2 + i) \) defined by: \[ z_{1} \otimes z_{2} = z_{1} + z_{2} + |z_{1} \cdot z_{2}| \] let's follow the steps below. ### Step 1: Add the Complex Numbers First, add the two complex numbers: \[ z_{1} + z_{2} = (1 - 2i) + (2 + i) = (1 + 2) + (-2i + i) = 3 - i \] ### Step 2: Multiply the Complex Numbers Next, multiply the two complex numbers: \[ z_{1} \cdot z_{2} = (1 - 2i)(2 + i) \] Expanding this: \[ = 1 \cdot 2 + 1 \cdot i + (-2i) \cdot 2 + (-2i) \cdot i \] \[ = 2 + i - 4i - 2i^2 \] \[ = 2 - 3i - 2(-1) \quad \text{(since } i^2 = -1\text{)} \] \[ = 2 - 3i + 2 \] \[ = 4 - 3i \] ### Step 3: Compute the Magnitude Now, find the magnitude of the product: \[ |z_{1} \cdot z_{2}| = |4 - 3i| = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \] ### Step 4: Combine the Results Finally, combine the results from Steps 1 and 3: \[ z_{1} \otimes z_{2} = (3 - i) + 5 = 8 - i \] ### Conclusion The result of the operation \( (1 - 2i) \otimes (2 + i) \) is: \[ 8 - i \] **Answer:** \( 8 - i \)

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Bonus Knowledge

To find the equivalent of the operation \( (1-2i) \otimes (2+i) \), we need to plug in the values into the defined operation. First, compute \( z_1 + z_2 \): \[ (1 - 2i) + (2 + i) = (1 + 2) + (-2i + i) = 3 - i \] Next, compute \( |z_1 \cdot z_2| \): \[ z_1 \cdot z_2 = (1 - 2i)(2 + i) = 1 \cdot 2 + 1 \cdot i - 2i \cdot 2 - 2i \cdot i = 2 + i - 4i + 2 = 4 - 3i \] Now, find the absolute value: \[ |4 - 3i| = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \] Now combine these results: \[ (1 - 2i) \otimes (2 + i) = (3 - i) + 5 = 8 - i \] Thus, the equivalent of the operation \( (1-2i) \otimes (2+i) \) is \( 8 - i \).

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