Question
Question: The real and imaginary parts of \[\dfrac{{{\text{a + ib}}}}{{{\text{a - ib}}}}\] are A. \({{\text{...
The real and imaginary parts of a - iba + ib are
A. a2 - b2, 2ab
B. a2 - b2a2 + b2, a2 - b22ab
C. a2 + b2a2 - b2, a2 + b22ab
D. a2 - b2a2 + b2, a2 + b22ab
Solution
Hint: To solve this question we will simplify the given complex number by multiplying it with its conjugate and then find the real and imaginary part of the complex number.
Complete step-by-step solution:
Now, every complex number has both real and imaginary parts. Complex numbers are written as z = x + iy, where x is the real part and y is the imaginary part, i is called iota having value of −1 i.e. i = −1 . The given complex number has both real and imaginary parts in both numerator and denominator, so we will simplify the complex number by multiplying it by the conjugate of the conjugate number. Conjugate of a complex number z is represented as {\text{\bar z}}. We can find it by changing the sign of the coefficient of iota(i). Now, we have to find the conjugate of the denominator because complex numbers have to be represented in the form with no term containing i in the denominator.
Let z = a - iba + ib
Conjugate of a – ib = a + ib
Now, multiply and divide the complex number with the conjugate, we get
z = (a - iba + ib)(a + iba + ib)
Solving z by using the property (a + ib)(a - ib) = a2 - b2 in the denominator.
z = ((a2 - (ib)2)(a + ib)2)
z = ((a2 + b2)(a2 + (ib)2 + i2ab)
z = (a2 + b2a2 - b2 + i2ab)
z = a2 + b2a2 - b2 + ia2 + b22ab
So, the real part of complex number is a2 + b2a2 - b2 and imaginary part of complex number is a2 + b22ab .
So, the answer is option (C).
Note: While solving such types of questions, we will first simplify the given complex number. First, we will check whether there is a term containing i in the denominator. If there is such a term, then we will multiply and divide the complex number by the conjugate of the denominator term. Then we will simplify the complex number as much as possible by using various properties. After simplification we can solve the given problem easily.