What is the answer to z^{-1}= 1/(3+2i) ?

The answer to z^{-1}= 1/(3+2i) is z=3+2i

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z^{-1}= 1/(3+2i)

z^{- 1} = \frac{1}{3 + 2 i}

z = 3 + 2 i

Solution steps

z^{- 1} = \frac{1}{3 + 2 i}

Substitute

z = x + y i

(x + y i)^{- 1} = \frac{1}{3 + 2 i}

Multiply both sides by

3 + 2 i

(x + y i)^{- 1} (3 + 2 i) = \frac{1}{3 + 2 i} (3 + 2 i)

Simplify

\frac{3 + 2 i}{x + i y} = 1

Multiply both sides by

x + i y

\frac{3 + 2 i}{x + i y} (x + i y) = 1 \cdot (x + i y)

Simplify

3 + 2 i = x + i y

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[3 = x 2 = y]

[3 = x 2 = y] : x = 3, y = 2

Substitute back

z = x + y i

z = 3 + 2 i

z^{2} - 2 i z - 3 = 0

x + i y = 0

(x + 2) + (2 y - 1) i = - 1 + 5 i

z^{2} = 1 + i

z - (\frac{1}{z}) = 3 + 2 i

What is the answer to z^{-1}= 1/(3+2i) ?
The answer to z^{-1}= 1/(3+2i) is z=3+2i