How can I proof that: $$(z^n)^* = (z^*)^n$$
Where: z is a complex number, n is a positive whole number * is the complex conjugate
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2Do you know how to prove that $(zw)^=z^w^*$? if so, just use induction. – Dennis Gulko Oct 13 '13 at 10:01
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Write $z = re^{i\theta}$, then $\overline{z} = re^{-i\theta}$. So $$ \overline{z}^n = r^n e^{-in\theta} = \overline{r^ne^{in\theta}} = \overline{z^n} $$
Prahlad Vaidyanathan
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Use induction. First check it for $n=2$: $$ (z^{\ast})^{2} = (a - ib)^{2} = a^{2} - 2i ab - b^{2} = (a^{2} + 2i ab - b^{2})^{\ast} = (z^{2})^{\ast} $$ Now, show that if it's true for every natural up to a certain $n$, then it's true for $n+1$.
ulilaka
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What do you plan to do with (z^2.z)* without assuming that it's equal to (z^2).z (which is the behaviour that we're trying to prove.) – Brondahl Apr 06 '22 at 13:02
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Alternatively, if you're happy to assume (zw)* = zw, then you don't need a complicated first step for n=2 – Brondahl Apr 06 '22 at 13:04
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For the next step in the proof by induction. Show that $x^*y^*=(xy)^*$ for all complex numbers. That allows you to then show that if $(x^*)^2=(x^2)^*$ then $(x^*)^2 \cdot x^* = (x^3)^*$, and so on.
K.defaoite
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David B
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As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. – Community Jun 13 '22 at 11:24