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

1. TuringTest

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

2. anonymous

how can you show me? please

3. 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?

4. TuringTest

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

5. anonymous

Thanks