Find the reciprocal of the complex number 3 + 2i Give your answer in the usual bi form?

Answer 1

This can be worked out in a number of ways:

First way

1/(3+2i) = x+yi

So 1 = (3+2i)(x+yi) = 3x +3yi + 2ix -2y = (3x-2y) + i(3y+2x)

As 1 = 1+0i

1 + 0i = (3x-2y) + i(3y+2x)

So we have two simultaneous equations:

3x - 2y = 1 ............. (1)
2x + 3y = 0 ............. (2)

Multiply (1) by 3 and (2) by 2 then add

9x - 6y = 3 ............. (1)*3
4x + 6y = 0 ............. (2)*2

13x = 3

x = 3/13 and y = (3x-1)/2 = (9/13 - 1)/2 = -2/13

Checking answers:

3x - 2y = 3(3/13) - 2(-2/13) = 9/13 + 4/13 = 1
2x + 3y = 2(3/13) + 3(-2/13) = 6/13 - 6/13 = 0

So 1/(3+2i) = (3-2i)/13 = 3/13 - (2/13)i

Which is the same as (3+2i)*/||(3+2i)||² = (3-2i)/(3²+2²) = (3-2i)/13

Second way is to express (3+2i) in polar form

3+2i = r(cosφ + isinφ)

And 1/(3+2i) = (3+2i)ֿ¹ = rֿ¹(cosφ + isinφ)ֿ¹ = (cosφ + isinφ)ֿ¹/r²

And as (cosφ + isinφ)ֿ¹ = (cosφ - isinφ)

1/(3+2i) = (3+2i)ֿ¹ = (cosφ - isinφ)/||(3+2i)||²

1/(3+2i) = (3-2i)/13 = 3/13 - (2/13)i

Answer 2

The reciprocal of 3 + 2i is...

(3-2i)/13 or (3/13) - (2i/13)

First you must rationalize the denominator by multiplying both the numerator and denominator by the conjugate of 3 + 2i. The conjugate of 3+ 2i is 3- 2i. multiplyin the numerators together you get 3 - 2i and multiplying the denominators together you get 9 - 4i^2. Since i^2 equals -1, the denominator changes to 9 + 4, which simplifies to 13. Therefore, the reciprocal of 3 + 2i is (3-2i)/13 OR (3/13) - (2i/13).

Answer 3

reciprocal is 1/(3+2i), but it is not in standard form
first of all try to rationalize the denominator by multiplying it with the conjugate i.e. 3-2i which gives u( 3-2i)/(3^2+2^2)
which finally gives u [(3/13)-i(2/13)] as the result

Answer 4

The reciprocal of (3+2i) is (3+2i)*(3-2i)/(3-2i) which on simplifying gives (3/13)-(2/13)i.

Answer 5

1/(3+2i) Multiply top and bottom by 3-2i

For the deonominator
(3+2i)(3-2i)=9-4i^2=9+4=13

So you get (3-2i)/13

In standard form divide the 13 into each term that is in the numerator

(3/13) + (-2/13) i

Answer 6

1/(3 + 2i )

Conjugate of 3+2i = 3-2i

Multiply both numerator and denominator by conjugate of 3+2i

= (3-2i)/(3+2i)(3-2i)

= (3-2i)/(3*3 -3*2i + 3*2i -4*i²)

= (3-2i)/(9 - 4*(-1))

= (3-2i)/(9+4)

= (3-2i)/13

= (3/13) - (2/13)i

Thats in a + bi form you needed!

Answer 7

1/(3+2i) multiply top and bottom by 3-2i
1/(3+2i) (3-2i)/(3-2i) = (3-2i)/(9+4) = (3-2i)/13
= (3/13) + (-2/13) i .

Answer 8

= 1/(3+2i)

= 1 * (3-2i)
------------
(3+2i) (3-2i)
= (3-2i)
-------------
(9 - 4i^2) i^2 = -1

= (3-2i) = (3-2i)
---------- ----------
9 - (-4) .. 13

Answer 9

1/[3+2i]
=[3-2i]/[3+2i][3-2i]
=[3-2i]/[9+4]
=3/13-[2/13]i
=.23-.15i