$A r g (z) = - A r g (z^{- 1})$ $Arg(z) = -Arg(z^-1)$ for any $z \in C, z \neq 0$ $zinC, z!=0$ How to prove it?

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$A r g (z) = - A r g (z^{- 1})$ $Arg(z) = -Arg(z^-1)$ for any $z \in C, z \neq 0$ $zinC, z!=0$ How to prove it?

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Answer 1 · 2017-06-11T09:25:28

Please see below.

Explanation:

Let $z = a + i b = r cos θ + i r sin θ$ $z=a+ib=rcostheta+irsintheta$

then $| z | = \sqrt{a^{2} + b^{2}}$ $|z|=sqrt(a^2+b^2)$ and $A r g (z) = {tan}^{- 1} (\frac{b}{a})$ $Arg(z)=tan^(-1)(b/a)$

Let $\frac{b}{a} = tan α$ $b/a=tanalpha$ then $A r g z = α$ $Argz=alpha$

Note that $z \neq 0 \Rightarrow$ $z!=0=>$ that both $a$ $a$ and $b$ $b$ cannot be together zero i.e. $a^{2} + b^{2} \neq 0$ $a^2+b^2!=0$

As such $z^{- 1} = \frac{1}{a + i b} = \frac{a - i b}{(a + i b) (a - i b)}$ $z^(-1)=1/(a+ib)=(a-ib)/((a+ib)(a-ib))$

= $\frac{a - i b}{a^{2} + b^{2}} = \frac{a}{a^{2} + b^{2}} - i \frac{b}{a^{2} + b^{2}}$ $(a-ib)/(a^2+b^2)=a/(a^2+b^2)-ib/(a^2+b^2)$

and $A r g (z^{- 1}) = {tan}^{- 1} ⎛ ⎝ \frac{- \frac{b}{a^{2} + b^{2}}}{\frac{a}{a^{2} + b^{2}}} ⎞ ⎠ = {tan}^{- 1} (- \frac{b}{a})$ $Arg(z^(-1))=tan^(-1)((-b/(a^2+b^2))/(a/(a^2+b^2)))=tan^(-1)(-b/a)$

= ${tan}^{- 1} (- tan α)$ $tan^(-1)(-tanalpha)$

But as $tan (- α) = - tan α$ $tan(-alpha)=-tanalpha$

$A r g (z^{- 1}) = - α = - A r g (z)$ $Arg(z^(-1))=-alpha=-Arg(z)$

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$A r g (z) = - A r g (z^{- 1})$ $Arg(z) = -Arg(z^-1)$ for any $z \in C, z \neq 0$ $zinC, z!=0$ How to prove it?

1 Answer

Explanation:

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$A r g (z) = - A r g (z^{- 1})$ $Arg(z) = -Arg(z^-1)$ for any $z \in C, z \neq 0$ $zinC, z!=0$ How to prove it?