howto:hambasics:sections:test
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howto:hambasics:sections:test [2021/02/13 19:04] – va7fi | howto:hambasics:sections:test [2021/02/13 19:14] (current) – [Cartesian vs Polar] va7fi | ||
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{{ complexroots.png? | {{ complexroots.png? | ||
- | So \$z = 2\$ was to be expected since \$ 2^3 = 8 \$ but it looks like there' | + | So \$z = 2\$ was to be expected since \$ 2^3 = 8 \$ but it looks like there are two more solutions. \\ \\ |
To find them, we first notice that the three solutions are spread out evenly around the circle, that is they are 120° apart. | To find them, we first notice that the three solutions are spread out evenly around the circle, that is they are 120° apart. | ||
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- | The lesson here is that since the polar representation uses exponents, and exponents turn multiplication into addition((\$ x^a \cdot x^b = x^{a+b} \$)), the polar representation is easiest for multiplication, | + | The lesson here is that since the polar representation uses exponents, and exponents turn multiplication into addition((\$ x^a \cdot x^b = x^{a+b} \$)), the polar representation is easiest for multiplication, |
===== Important Algebraic Results ===== | ===== Important Algebraic Results ===== | ||
* Use the Euler identity to get the following two useful results: | * Use the Euler identity to get the following two useful results: |
howto/hambasics/sections/test.1613271888.txt.gz · Last modified: 2021/02/13 19:04 by va7fi