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.

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

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

$=( 10 + 25i -4i - 10i^2)/(25 -4i^2) =( 10 + 21i + 10)/29$

$=( 20 + 21i)/29$

$rArr( 2 + 5i)/(5 + 2i ) = 20/29 + 21/29 i$

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