Which pair of complex numbers has a real-number product? (1 + 2i)(8i) (1 + 2i)(2 – 5i) (1 + 2i)(1 – 2i) (1 + 2i)(4i)
2 answers:
Answer:
(1+2i)(1-2i)
Step-by-step explanation:
Following are the pairs of the complex number:
(1+2i)(8i),
(1 + 2i)(2 – 5i)
(1+2i)(1-2i) and (1+2i)(4i)
We have to check which pair out of these is a real number product, which means which pair do not contain terms consisting of "i".
A.
=
B.
=
C.
=
D.
=
Since, A,B,D contains the term "i" which means they are not real valued, therefore option C that is (1+2i)(1-2i) has a real number product.
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Answer:
(-2, 4, -2)
x=-2, y=4, z=-2.
Step-by-step explanation:
So we have the three equations:
And we want to find the value of each variable.
To solve this system, first look at it and consider what you should try to do.
So we can see that the second and third equations both have an x.
Therefore, we can isolate the variables for the second and third equation and then substitute them into the first equation to make the first equation all xs.
Therefore, let's first isolate the variable in the second and third equation.
Second Equation:
First, divide everything by -2 to simplify things:
Subtract x from both sides. The xs on the left cancel:
Now, divide everything by -2 to isolate the z:
So we've isolated the z variable. Now, do the same to the y variable in the third equation:
Subtract x from both sides:
Divide both sides by 2:
Now that we've isolated the y and z variables, plug them back into the first equation. Therefore:
Distribute the third term. The -2s cancel out:
Since there is still a fraction, multiply everything by 2 to remove it:
Distribute:
Combine like terms:
Add 2 to both sides:
Divide both sides by 7:
Therefore, x is -2.
Now, plug this back into the second and third simplified equations to get the other values.
Second equation:
Third equation:
Therefore, the solution is (-2, 4, -2)
