Algebra 2: What is the multiplicative inverse of 1+i

..1........1 - i
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1+ i......1 - i

1 - i
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1 - i^2

1 - i
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1 - (-1)

1 - i
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1 + 1

1 - i
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..2

1.........1
***..-..*** i ...........ANSWER
2.........2
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To find the multiplicative inverse of any non-0 complex number a + bi,

solve the equation (a + bi)(x + yi) = 1


This complex equation yields 2 real equations:

ax -- by = 1

bx + ay = 0

The determinant = a^2 + b^2, which is > 0 because at least one of a and b is non-zero.
Therefore this pair of equations has a unique solution.

For the complex number 1 + i , a = b = 1

Then

x - y = 1

x + y = 0

Solution: x = 1/2, y = - 1/2

x + iy = 1/2 - i/2 = (1 - i)/2

A multiplicative inverse is a number wherein when you multiply the given number with its multiplicative inverse, you get "1".

So to do this, you must divide 1 by the given number to get its multiplicative inverse. Thus,

1/(1+i)

Then multiply the numerator and denominator by the conjugate of 1+i which is 1-i. you'll get the following.

(1+i)/1+1 = (1+i)/2..

Tell your professor, "Who's your daddy now?"

=1/(1+i) = (1-i) / [ (1+i)(1-i) ] = (1-i) / [1-i^2] = (1-i)/ [ 1-(-1) ] = (1/2) (1-i)

-1-i . you have to multiply the expression by -1.

(1/2) - (1/2)i

1-i