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What would be the complex conjugate for these three. Assuming $i$ is always $${\sqrt{-1}}$$

$$i^{11}$$ $$(2-3i)^3$$ $$\frac{3-i}{2i+5}$$

Gerry Myerson
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user1730308
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    Step one: write each expression in the form a+bi with a,b real. For the first, use i^4=1. For the second, simply expand ("foil") and collect terms. For the third, multiply numerator and denominator by the conjugate of the denominator. These are standard methods. – anon Mar 24 '13 at 22:48
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    Complex conjugation respects multiplication and addition. So the answers are $(-i)^{11}$, $(2+3i)^3$, $(3+i)/(-2i+5)$. – Cocopuffs Mar 24 '13 at 22:52
  • Better yet, the answers are $\overline{i^{11}}$, $\overline{(2-3i)^3}$, and $\overline{(3-i)/(2i+5)}$. – Sean Eberhard Mar 26 '13 at 12:42
  • Just adding a pedantic approach to Cocopuffs' remark: $\phi(z):=\overline{z}$ is a real algebra automorphism of $\mathbb{C}$ (that's even an involution, meaning it is equal to its own inverse). That is $\phi(sz_1+tz_2)=s\phi(z_1)+t\phi(z_2)$ and $\phi(z_1z_2)=\phi(z_1)\phi(z_2)$ for every $s,t\in\mathbb{R}$ and every $z_1,z_2\in\mathbb{C}$, and $\phi(1)=1$. Note this also implies $\phi(1/z)=1/\phi(z)$ for every $z\neq 0$. – Julien Mar 26 '13 at 12:43

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$$i^{11}=-i \Longrightarrow -i^*=i$$ $$(2-3i)^3=-46-9i\Longrightarrow (-46-9i)^*=-46+9i$$ $$\frac{3-i}{2i+5}=\frac{13}{29}-\frac{11}{29}i\Longrightarrow\left(\frac{13}{29}-\frac{11}{29}i\right)^*=\frac{13}{29}+\frac{11}{29}i$$

Jan Eerland
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