How do you divide #(2+5i)/(5+2i)#?
2 Answers
Explanation:
To divide
This way, you will be able to "get rid" of the
Let me walk you through the process:
Your denominator is
To extend the fraction, you need to multiply both the numerator and the denominator with
#(2+5i)/(5+2i) = ((2 + 5i) * (5-2i)) / ((5+2i) * (5 - 2i)) = (10 + 25 i - 4 i - 10 i ^2) / (5^2 - (2i)^2) = (10 + 21 i - 10 i^2) / (25 - 4i^2)#
...remember that
# = (10 + 21i + 10) / (25 + 4) = (20 + 21i) / 29 = 20/ 29 + 21/29 i#
Explanation:
Multiply the numerator and denominator by the complex conjugate of 5 + 2i . This ensures that the denominator is real.
The complex conjugate of a complex number a + bi is a - bi.
note that (a + bi )(a - bi ) =
#( i = sqrt- 1 rArri^2 = (sqrt- 1 )^2 = - 1 ) #
#=( 10 + 25i -4i - 10i^2)/(25 -4i^2) =( 10 + 21i + 10)/29 #
#=( 20 + 21i)/29 #
#rArr( 2 + 5i)/(5 + 2i ) = 20/29 + 21/29 i #