How do you divide #(2+5i)/(5+2i)#?

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Answer 1 · 2016-01-13T10:04:18

#20/ 29 + 21/29 i#

To divide #(2+5i)/(5+2i)#, you need to find the complex conjugate of the denominator and extend the fraction with it.

This way, you will be able to "get rid" of the #i# in the denominator.

Let me walk you through the process:

Your denominator is #5 + 2i#, thus the complex conjugate is # 5 - 2i#.

To extend the fraction, you need to multiply both the numerator and the denominator with #5 - 2i#:

#(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 #i^2 = -1# ...

# = (10 + 21i + 10) / (25 + 4) = (20 + 21i) / 29 = 20/ 29 + 21/29 i#

Answer 2 · 2016-01-13T10:27:46

# 20/29 +21/29 i #

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 ) = #a^2 + abi - abi - bi^2 = a^2 + b^2 # which is real.

#( i = sqrt- 1 rArri^2 = (sqrt- 1 )^2 = - 1 ) #

#rArr ((2 +5i )(5 - 2i))/((5 + 2i)(5 - 2i) #

#=( 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 #