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frx

  • 3 years ago

Show that the equation \[z ^{3}+(-3+i)z^{2}+(6-3i)z+(-6-8i)=0\] has a pure imaginary root and then solve the equation. I know how to solve it by setting z=bi but what if the description would have said "has a pure real root " should i use z=a instead?

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  1. hartnn
    • 3 years ago
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    yes where a is real.

  2. frx
    • 3 years ago
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    Thank you, that was all :)

  3. hartnn
    • 3 years ago
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    quickest post ever! welcome ^_^

  4. frx
    • 3 years ago
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    Haha yeah that troubled me a bit since I always deal with pure imaginary ones :p I assume it's the same if it doesn't have any pure root then it's z=a+bi, right?:)

  5. hartnn
    • 3 years ago
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    right again :)

  6. frx
    • 3 years ago
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    Great, thanks!

  7. hartnn
    • 3 years ago
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    welcome ^_^

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