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If $w$ and $z$ are two complex number, prove that $$\left(\frac{w}{z}\right)^* = \frac{w^*}{z^*}$$ My attempt:$$\left(\frac{w}{z}\right)^* = \left(w \cdot \frac{1}{z}\right)^*$$ $$\left(w \cdot \frac{1}{z}\right)^* = w^* \cdot \left(\frac{1}{z}\right)^*$$ Hence, $$\left(\frac{w}{z}\right)^* = \frac{w^*}{z^*}$$

Anne Bauval
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James Thomson
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  • If to you $\frac1{z^}=\left(\frac1z\right)^$ is a given, then you're done. –  Aug 13 '23 at 16:03
  • I'd say write them in polar form and it'd be neater, but like Dumper said if the complex conjugate of an inverse is the inverse of the conjugate then this is fine – Lourenco Entrudo Aug 13 '23 at 16:26
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    It depends what you are supposed to know, i.e. how $*$ was defined to you. Was it only in cartesian coordinates? or polar as well? – Anne Bauval Aug 13 '23 at 16:44
  • @LourencoEntrudo Or just use the formulas for real and imaginary part, because the sooner you learn them by heart the better. –  Aug 13 '23 at 16:45
  • @DumperDGarb I suggested polar form because the division in standard real and imaginary part may be less straightforward to apply the definition of the conjugate – Lourenco Entrudo Aug 13 '23 at 17:47
  • @LourencoEntrudo Very arguably, but that was what "the sooner the better" stood for. –  Aug 13 '23 at 17:49

2 Answers2

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Hint: We can consider real and imaginary part of $w$ and $z$ in the form \begin{align*} w&=a+ib\qquad\qquad a,b\in\mathbb{R}\\ z&=c+id\qquad\qquad c,d\in\mathbb{R} \end{align*}

We obtain by consequently splitting the complex number into real and imaginary part \begin{align*} \color{blue}{\left(\frac{w}{z}\right)^{*}}&=\left(\frac{a+ib}{c+id}\right)^{*}\\ &=\left(\frac{(a+ib)(c-id)}{c^2+d^2}\right)^{*}\\ &=\left(\frac{ac+bd}{c^2+d^2}-i\frac{ad-bc}{c^2+d^2}\right)^{*}\\ &\color{blue}{=\frac{ac+bd}{c^2+d^2}+i\frac{ad-bc}{c^2+d^2}} \end{align*} On the other hand we get \begin{align*} \color{blue}{\frac{w^{*}}{z^{*}}}&=\frac{a-ib}{c-id}=\cdots \end{align*}

Can you continue?

Markus Scheuer
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I think what you are looking for is as follows: Let $w=ae^{i\alpha}$ and $z=be^{i\beta}$. Then

$$\bigg(\frac{w}{z}\bigg)=\frac{a}{b}e^{i(\alpha-\beta)}$$

So,

$$\bigg(\frac{w}{z}\bigg)^*=\frac{a}{b}e^{-i(\alpha-\beta)}=\frac{ae^{-i\alpha}}{be^{-i\beta}}=\frac{w^*}{z^*}$$

Obviously, this is general and applicable to any values of the parameters.

Cye Waldman
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