When multiplying two complex numbers, the imaginary parts cancel each other out, resulting in a real number without an imaginary component. This can be illustrated through examples, such as (3 - 2i) and (3 + 2i) simplifying to 13 and (2 + 5i) and (2 - 5i) simplifying to 29.
When multiplying two complex numbers, the product can simplify to an integer if the imaginary parts cancel each other out. For example, let's consider the product of (3 - 2i) and (3 + 2i):
(3−2i)(3+2i) = 9−6i+6i−4i^2
Simplifying, we know that i^2=−1, so: 9−4(−1)=13
The result, 13, is a real number without an imaginary component.
Another example is the product of (2 + 5i) and (2 - 5i):
(2+5i)(2−5i) = 4−10i+10i−25i^2
Simplifying, i^2=−1, so: 4−25(−1)=29
In this case, the product also simplifies to an integer, 29, demonstrating the concept further.
However, there are cases where the product of two complex numbers does not simplify to an integer, and the imaginary part is retained, like the product of (5 - 2i) and (7i) resulting in a complex number (14 + 35i).
The probable question may be:
In the realm of complex numbers, elucidate how the product of two complex numbers can simplify to an integer, thereby eliminating the imaginary component (i). Support your explanation with a concrete example.
Explanation: When multiplying two complex numbers, the product can simplify to an integer if the imaginary parts cancel each other out. Consider the product of (3 - 2i) and (3 + 2i):
(3−2i)(3+2i)=9−6i+6i−4i^2
Simplifying, we know that i^2=−1, so:
9−4(−1)=13
The result, 13, is a real number without an imaginary component.
Additional Information:
Consider the product of two other complex numbers, (2 + 5i) and (2 - 5i):
(2+5i)(2−5i)=4−10i+10i−25i^2
Simplifying, i^2=−1:
4−25(−1)=29
In this case, the product also simplifies to an integer, 29, demonstrating the concept further.
Counterexample:
As a counterexample, the product of (5 - 2i) and (7i) results in a complex number (14 + 35i), where the imaginary part is retained.