Let z= (7+2i)/(6-3i)
Let us multiply by 6+3i
=> z= (7+2i)(6+3i)/(6-3i)(6+3i)
= (42 + 21i + 12i - 6)/(6^2 + 3^2)
= (36 + 33i)/(45)
==> z= (36/45) + (33/45)i
= (4/5) + (11/25)i
==> z' = (4/5) - (11/25)i
The multiplicative inverse of a number X is a number Y such that X*Y=1.
Here X= (7+2i)/(6-3i)
Therefore the multiplicative inverse is (6-3i)/(7+2i).
The required multiplicative inverse is (36-33i)/53
We'll establish the multiplicative inverse of the given ratio is:
According to the rule, it is not allowed that the denominator to be a complex number.
We'll multiply the ratio by the conjugate of (7+2i).
The conjugate of 7+2i = 7-2i
(6-3i)/(7+2i) = (6-3i)(7-2i) / (7+2i)(7-2i)
We'll remove the brackets:
(6-3i)(7-2i) = 42 - 12i - 21i - 6
We'll combine the real parts and the imaginary parts and we'll get:
Now, we'll calculate the difference of squares:
(7+2i)(7-2i) = (7)^2 - (4i)^2 = 49 + 16 = 65
(6-3i)/(7+2i) = ( 36 - 33i)/65
The multiplicative inverse of the number (7+2i)/(6-3i) is:
36/65 - (33/65)*i
The multiplicative inverse of a number X is equal to `1/X` .
For `X = (7+2i)/(6-3i)` , the multiplicative inverse is `1/X = 1/((7+2i)/(6-3i))`
To simplify this multiply the numerator and denominator with the complex conjugate of the denominator.
= `((6-3i)*(7-2i))/(7^2 + 2^2)`
= `(42 - 21i - 12i + 6i^2)/(53)`
= `(42 - 21i - 12i - 6)/(53)`
= `(36 - 33i)/(53)`
To find the multiplicative inverse of (7+2i)(6-3i).
The multiplicative inverse of z = (a+bi) is z' and zz' = 1.
Or z' = 1/z.
First we simplify (7+2i)(6-3i) = 42 -21i+12i -6i^2
= 42-9i +6, as i^2 = -1
Therefore z = (48-9i
z' = 1/(48-9i). Realise the denomonator by multiplying (48+9i) both numerator and denominators.
z' = (48+9i)/(48^2+9^2)
z' = (48/2385) - (9/2385)i
z' = 16/795 -(3/795)i