Question

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

Answer 1

just use the quadratic formula
the only tricky part is taking the square root...

Answer 2

how can you show me? please

Answer 3

\[z=\frac{-(3+4i)\pm\sqrt{(3+4i)^2-4(-1+5i)}}2\]\[=\frac{-(3+4i)\pm\sqrt{9+24i-16+4-20i)}}2\]\[=\frac{-(3+4i)\pm\sqrt{-3+4i}}2\]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?

Answer 4

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

Answer 5

Thanks