anonymous
  • anonymous
How do I solve z^2+(3+4i)z-1+5i=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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TuringTest
  • TuringTest
just use the quadratic formula the only tricky part is taking the square root...
anonymous
  • anonymous
how can you show me? please
TuringTest
  • TuringTest
\[z=\frac{-(3+4i)\pm\sqrt{(3+4i)^2-4(-1+5i)}}2\]\[=\frac{-(3+4i)\pm\sqrt{9+24i-16+4-20i)}}2\]\[=\frac{-(3+4i)\pm\sqrt{-3+4i}}2\]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?

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

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