What is the multiplicative inverse of 6-3i?
The multiplicative inverse of a term x is given by y such that x*y = 1.
Now we have to find the multiplicative inverse of 6 - 3i
Let it be x + yi
(6 - 3i)(x + yi) = 1
=> x + yi = 1/(6 - 3i)
=> x + yi = (6 + 3i)/(6 - 3i)(6 + 3i)
=> x + yi = (6 + 3i)/ (6^2 + 3^2)
=> x + yi = (6+ 3i)( 36 + 9)
=> x + yi = (6 + 3i)/ 45
=> x + yi = 6/45 + (3/45)i
The multiplicative inverse is 6/45 + (3/45)i
The multiplicative inverse is the fraction whose numerator is 1 and denominator is the given complex number.
We'll note the inverse as x:
(6-3i)*x = 1
We'll divide by (6-3i) both sides:
x = 1/(6-3i)
Since it is not allowed to keep a complex number to denominator, we'll multiply the ratio by the conjugate of the complex number:
x = (6+3i)/(6-3i)(6+3i)
We'll re-write the denominator as a difference of squares:
(6-3i)(6+3i) = 6^2 - (3i)^2
(6-3i)(6+3i) = 36 - 9i^2
But i^2 = -1:
(6-3i)(6+3i) = 36 + 9
(6-3i)(6+3i) = 45
We'll re-write x:
x = (6+3i)/45
x = 6/45 + 3i/45
We'll simplify and we'll get:
x = 2/15 + i/15
The multiplicative inverse of the complex number 6 - 3i is 2/15 + i/15.