668 Worked-Out Solutions to Exercises: Chapters 21 to 29
Now we can make these substitutions:- Leta from the formula equal a in our problem
- Letb from the formula equal b in our problem
- Letc from the formula equal a in our problem
- Letd from the formula equal −b in our problem
The signs will be tricky, now! The quotient formula looks like this:(a+jb) / [a+j(−b)]
= (aa+b× (−b)] / [a^2 + (−b)^2 ]+j[ba−a× (−b)] / [a^2 + (−b)^2 ]For any real number b, (−b)^2 =b^2. Knowing that, and simplifying the above expression as
much as possible, we get(aa+b× (−b)] / [a^2 + (−b)^2 ]+j[ba−a× (−b)] / [a^2 + (−b)^2 ]
= (a^2 −b^2 ) / (a^2 +b^2 )+j[2ab / (a^2 +b^2 )]- To find (a−jb) / (a+jb), let’s first change the subtraction in the numerator to negative
addition. That will give us the expression
[a+j(−b)] / (a+jb)Now we can again use the quotient formula for complex numbers. This time, let’s make
these substitutions:- Leta from the formula equal a in our problem
- Letb from the formula equal −b in our problem
- Letc from the formula equal a in our problem
- Letd from the formula equal b in our problem
Once again, we must pay close attention to the signs. The quotient formula now looks
like this:[a+j(−b)] / (a+jb)
= (aa+ (−b)×b] / (a^2 +b^2 )+j(−ba−ab) / (a^2 +b^2 )Simplifying the above expression as much as possible, we get(aa+b× (−b)] / [a^2 + (−b)^2 ]+j[ba−a× (−b)] / [a^2 + (−b)^2 ]
= (a^2 −b^2 ) / (a^2 +b^2 )+j[− 2 ab / (a^2 +b^2 )]
= (a^2 −b^2 ) / (a^2 +b^2 )−j[2ab / (a^2 +b^2 )]This is the complex conjugate of the result we got in Prob. 8.- If k is a positive real number, then two pure real numbers have absolute values equal
tok. These numbers are k and −k. Two pure imaginary numbers, jk and −jk, also have