plz explain for me.?

Question

let z=r(cos(x)+i*sin(x))
why is 1/z=r(cos(x)-i*sin(x))

Answer 1

It's some trigonometry, but mostly it's complex analysis.

Also, I think there's a mistake. In the equation for 1/z, I think you want to replace r with 1/r, so that we want to show that the multiplicative inverse of z has the value

1/z = 1/r[cos(x)-i*sin(x)]

So you just need to show that if you multiply it times z, you'll get 1.

Show that 1 is equal to
z * 1/z = r*(cos(x)+i*sin(x)) * 1/r*(cos(x)-i*sin(x))

To do that, first rearrange to put 1/r next to r so they cancel out:
= r*1/r*(cosx+isinx)*(cosx-isinx)
= (cosx+isinx)*(cosx-isinx)

[For shorthand, I've written "cos(x)" as cosx and "i*sin(x)" as isinx.]

Use distributive property to multiply each term inside the parantheses by a term in the other parantheses:
= cosx*cosx - isinx*cosx + isinx*cosx - isinx*isinx

the middle two terms add up to zero (one is a negative of the other)
= cosx*cosx - isinx*isinx

In the second term, rearrange to put the two "i's" next to each other, which will multiply together to give -1 (by definition of "i"):
= cosx *cosx - i*i*sinx*sinx
= cosx*cosx + sinx*sinx

And now the trigonometry part:
(cosx)^2 + (sinx)^2 = 1

which is the result we were looking for, which means we've proved that
1/z = 1/r (cos(x) - i*sin(x)).

QED

Answer 2

z*1/z=1

Answer 3

I guess 1/z=(cos(x)-i*sin(x))/r and not 1/z=r(cos(x)-i*sin(x))
z x 1/z = 1 so
r(cos(x)+i*sin(x)) * (cosx - isinx)/r = 1 must be true for all x.
So (cos(x)+i*sin(x)) * (cosx - isinx) = 1 must be true for all x.
And really (cosx)^2 + (sinx)^2 = 1

Answer 4

It's a trigonometry question.