Compute Each Quotient

[nycmathguy]

David Cohen Textbook Questions.
Enjoy.

 

[Country Boy]

What is intended here is "rationalizing the denominator" so that the complex number can be written in the "standard form", a+ bi, with a and b real numbers. You can do that by multiplying both numerator and denominator by the "complex conjugate" of the denominator.

The complex conjugate of the complex number, a+ bi, is a- bi. The product of a complex number and its complex conjugate is a real number- (a+ bi)(a- bi)= a^2- abi+ abi- b^2i^2= a^2+ b^2.

Problem A is i/(5+ i). The complex conjugate of 5+ i is 5- i. i(5- i)= 5i- i^2= 1+ 5I. (5+ i)( 5- i)= 25- 5i+ 5i- i^2= 25+ 1= 26. i/(5+ i)= (1+ 5i)/26.

Problem B is (i+ i^2)/(i^3+ i^4). As you have written on the page, i^2= -1. So i^3= (i^2)i= -i and i^4= (i^2)(i^2)= (-1)(-1)= 1.(i+ i^2)/(i^3+ i^4)= (i-1)/(-i+ 1)= (-1+ i)/(1- i)= -(1- i)/(1- i)= -1.

 

[nycmathguy]

What is intended here is "rationalizing the denominator" so that the complex number can be written in the "standard form", a+ bi, with a and b real numbers. You can do that by multiplying both numerator and denominator by the "complex conjugate" of the denominator.

The complex conjugate of the complex number, a+ bi, is a- bi. The product of a complex number and its complex conjugate is a real number- (a+ bi)(a- bi)= a^2- abi+ abi- b^2i^2= a^2+ b^2.

Problem A is i/(5+ i). The complex conjugate of 5+ i is 5- i. i(5- i)= 5i- i^2= 1+ 5I. (5+ i)( 5- i)= 25- 5i+ 5i- i^2= 25+ 1= 26. i/(5+ i)= (1+ 5i)/26.

Problem B is (i+ i^2)/(i^3+ i^4). As you have written on the page, i^2= -1. So i^3= (i^2)i= -i and i^4= (i^2)(i^2)= (-1)(-1)= 1.(i+ i^2)/(i^3+ i^4)= (i-1)/(-i+ 1)= (-1+ i)/(1- i)= -(1- i)/(1- i)= -1.

Looks good.