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anonymous

  • 4 years ago

How do I solve z^2+(3+4i)z-1+5i=0

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  1. TuringTest
    • 4 years ago
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    just use the quadratic formula the only tricky part is taking the square root...

  2. anonymous
    • 4 years ago
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    how can you show me? please

  3. TuringTest
    • 4 years ago
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    \[z=\frac{-(3+4i)\pm\sqrt{(3+4i)^2-4(-1+5i)}}2\]\[=\frac{-(3+4i)\pm\sqrt{9+24i-16+4-20i)}}2\]\[=\frac{-(3+4i)\pm\sqrt{-3+4i}}2\]the technique for the square root here is long winded, but I just considered that the the numbers must foil out to that expression and saw that \[\sqrt{-3+4i}=1+2i\]because\[(1+2i)(1+2i)=-3+4i\]is that at all obvious to you? or do you want to see the whole procedure?

  4. TuringTest
    • 4 years ago
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    so you get\[z=\frac{-(3+4i)\pm(1+2i)}2=\left\{ -2-3i,-1-i \right\}\]as a final answer

  5. anonymous
    • 4 years ago
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    Thanks

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