Multiplicative inverse of complex numbers

[sarah786]

I can't find a proof for the multiplicative inverse of complex numbers... can anybody please tell me the proof (i already know what the formula is)

 

[micromass]

If you already know the formule than you're already on the good way. So, I guess the formula you have is $$z^{-1}=\frac{\overline{z}}{|z|^2}$$.

So the only thing you need to show now is that $$zz^{-1}=z^{-1}z=1$$. Just complete the following multiplication:

$$zz^{-1}=\frac{z\overline{z}}{|z|^2}=...$$

 

[LCKurtz]

If you already know the formule than you're already on the good way. So, I guess the formula you have is $$z^{-1}=\frac{\overline{z}}{|z|^2}$$.

So the only thing you need to show now is that $$zz^{-1}=z^{-1}z=1$$. Just complete the following multiplication:

$$zz^{-1}=\frac{z\overline{z}}{|z|^2}=...$$

I'll add to what Micromass has said. If you want the inverse of z = a+bi you are looking for a complex number w = x+yi such that

(a+bi)(x+yi) = 1 = 1+0i

Multiplying out the left side:

(ax - by) + (bx + ay)i = 1 + 0i

Equating real and imaginary parts:

ax - by = 1
bx + ay= 0

Solving these for x and y by determinants gives:

$$x = \frac{\left|\begin{array}{cc} 1 & -b\\0 & a \end{array}\right|} {\left|\begin{array}{cc} a & -b\\b & a \end{array}\right|} = \frac{a}{a^2+b^2},\, y = \frac{\left|\begin{array}{cc} a & 1\\b & 0 \end{array}\right|} {\left|\begin{array}{cc} a & -b\\b & a \end{array}\right|} = \frac{-b}{a^2+b^2}$$

This tells you that the inverse of z is

$$w = \frac{a}{a^2+b^2} + \frac{-b}{a^2+b^2}i = \frac{1}{a^2+b^2}(a-bi)=\frac{\overline z}{|z|^2}$$

 

[HallsofIvy]

Or, just to put in my oar, to find the multiplicative inverse of a+ bi, write
[tex]\frac{1}{a+ bi}[/itex]
and "rationlize the denominator". Multiply both numerator and denominator by a- bi:
[tex]\frac{1}{a+ bi}\frac{a- bi}{a- bi}= \frac{a- bi}{a^2+ b^2}[/itex]
which is, of course, exactly micromass's
$$\frac{\overline{z}}{|z|^2}$$.

That is the formula, which you say you already know. The "proof" (that that formula is correct) is just to multiply:
$$(a+ bi)\frac{a- bi}{a^2+ b^2}= \frac{(a+ bi)(a- bi)}{a^2+ b^2}= \frac{a^2+ b^2}{a^2+ b^2}= 1$$

 

[CellCoree]

The multiplicative inverse of a complex number is defined as a number that, when multiplied by the original complex number, results in a product of 1. In other words, it is the reciprocal of the complex number. To prove this, we can use the basic properties of complex numbers.

Let z = a + bi be a complex number, where a and b are real numbers and i is the imaginary unit.

We want to find a complex number w = c + di such that zw = 1.

Expanding the product zw, we get:

zw = (a + bi)(c + di)

= ac + adi + bci + bdi^2 (using the distributive property)

= ac + adi + bci - bd (since i^2 = -1)

= (ac - bd) + (ad + bc)i

We want this to be equal to 1, so we can equate the real and imaginary parts separately:

ac - bd = 1 (1)

ad + bc = 0 (2)

Solving for c and d in terms of a and b:

c = 1/a (3)

d = -b/a (4)

Substituting (3) and (4) into (2), we get:

a(-b/a) + b(1/a) = 0

-b + b = 0

Therefore, (3) and (4) satisfy (2). This means that w = c + di = 1/a - b/a i is the multiplicative inverse of z = a + bi.

To summarize, the proof for the multiplicative inverse of complex numbers is based on the properties of complex numbers and the algebraic manipulation of the original definition. The resulting inverse is a complex number that satisfies the condition of producing a product of 1 when multiplied by the original complex number.