We have to find the multiplicative inverse of 4 - 2i.
Now the multiplicative inverse of any number X is defined as X^-1, so that X*X^-1 = 1.
For 4 - 2i, let the multiplicative inverse be M.
Now M* (4 - 2i) = 1
=> M = 1/ (4 -2i)
=> M = (4 +2i)/ ( 4- 2i)( 4 +2i)
=> M = (4 + 2i) / [4^2 - (2i)^2]
=> M = ( 4 + 2i) / [ 16 + 4]
=> M = (4 + 2i) / 20
=> M = 4/20 + 2i /20
=> M = 1/5 + i/10
Therefore the multiplicative inverse of 4 - 2i is 1/5 + i/10.
To find the multiplicative inverse of the complex number.
We know the multiplication identity element of the complex number (x+yi) = e(x+iy) .
Therefore xe = x and and xi*e = xi. Therefore e = 1.
Therfore if the multiplicative inverse of 4-2i is x+yi, then
(4-2i)(x+yi) = 1.
4x+4yi-2xi -2yi^2 = 1+0*i
4x +(4y-2x)i +2y = 1+0*i.
(4x+2y) +(4y-2x)i = 1+0*i.
Equate imaginary parts and equate real parts both sides:
Imaginary parts: 4y -2x = 0.....(1)
Real Parts: 4x+2y = 1.....(2)
From (1): 4y- 2x = 0.
x= 2y. Substitute x= 2y in eq (2):
4(2y) +2y = 1
10y = 1.
y = 1/10.
x =2y= 2/10.
Therefore (4-2i) has the multiplicative inverse 2/10+i/10.
4y + 2(2y) =
The multiplicative inverse is get when we multiply the given complex number by the inverse number we'll get the result 1.
We'll note the inverse as x:
(4-2i)*x = 1
We'll divide by (4-2i) both sides:
x = 1/(4-2i)
We are not allowed to keep a complex number to denominator, so we'll multiply the ratio by the conjugate of the complex number:
x = (4+2i)/(4-2i)(4+2i)
We'll re-write the denominator as a difference of squares:
(4-2i)(4+2i) = 4^2 - (2i)^2
(4-2i)(4+2i) = 16 - 4i^2
But i^2 = -1:
(4-2i)(4+2i) = 16 + 4
(4-2i)(4+2i) = 20
We'll re-write x:
x = (4+2i)/20
x = 4/20 + 2i/20
We'll simplify and we'll get:
x = 1/5 + i/10
The multiplicative inverse of the complex number 4 - 2i is 1/5 + i/10.