Question
Question: Solving \(3-2yi={{9}^{x}}-6i\), for \(x\) and \(y\) real, we get: 1\. \(x=0.5,y=3.5\) 2\. \(x=0....
Solving 3−2yi=9x−6i, for x and y real, we get:
1. x=0.5,y=3.5
2. x=0.5,y=3
3. x=0.5,y=7
4. None of these
Solution
To solve the given question in which we have an imaginary equation as 3−2yi=9x−6i, we need to compare the real number with real number and coefficient of imaginary number to coefficient of imaginary number. After that, we will do required mathematical calculations to simplify the obtained expression and will find the required values.
Complete step by step answer:
Since, we have the given expression from the question as:
⇒3−2yi=9x−6i
Now, we will compare the corresponding values and variables of the left hand side to corresponding values and variables of the right hand side one by one. First of all, we will do the comparison for the real part of the expression as:
⇒3=9x
As we know, 3 is the square root of 9. So, we can write it in under root of 9 as:
⇒9=9x
Here, we can write the under root in the form of power as 21. So, we can write the above step as:
⇒921=9x
Now, the base will be canceled because the base is the same on both sides. So, we will have the above expression as:
⇒x=21
Here, we will convert the fraction into decimal as:
⇒x=0.5
Since, we got the value of x. Now, we will do the comparison of imaginary part of the expression as:
⇒2y=6
Here, we will divide by 2 each side of the above step as:
⇒22y=26
After solving it, we will get the value of y as:
⇒y=3
Hence, we will have the value of x and y as 0.5 and 3 respectively.
So, the correct answer is “Option 2”.
Note: A complex number is a number that contains both the real number and imaginary unit within itself. It is expressed as x+iy, where x and y are real numbers and iis an imaginary unit that satisfies the condition i2=−1.