Lorraine Harvey

2022-01-16

Show that additive inverse of any is unique and it equals
Show that multiplicative inverse of any non-zero is unique and it equals , where
$u=\left(\frac{x}{{x}^{2}+{y}^{2}}\right)$
$v=\frac{-y}{{x}^{2}+{y}^{2}}$

poleglit3

Expert

Step 1
We are aware that zero is the sum of any number and its additive inverse.
To find: Additive Inverse of the Complex Number:
Lets take the complex numbers $z=x+iy$ and its inverse be $-z=a+ib$
Step 2
Add the z and -z together to get
$z+\left(-z\right)=x+iy+\left(a+ib\right)=0$
Combine Like term, we get
$\left(x+a\right)+i\left(y+b\right)=0$
Using the zero product property, we can calculate an equivalent as
$a=-x$
$\left(x+a\right)=0$
and $b=-y$
$i\left(y+b\right)=0$
$a=-x$ and $b=-y$
Thus the additive inverse of the complex numbers z is $-z=-x+i\left(-y\right)$
Step 3
The multiplicative inverse of a complex number can be found by:
Lets take the complex number $z=x+iy$ and its

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