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Complex Numbers: solve (1-i)*z^2+(1+3i)z+(-4+2i)=0
- Thread starter DarkSun
- Start date
Re: Complex Numbers
There are two solutions, as you know.
\(\displaystyle z=\frac{-(1+3i)\pm\sqrt{(1+3i)^{2}-4(1-i)(-4+2i)}}{2(1-i)}\)
These whittle down to simple solutions. Nothing messy. \(\displaystyle \sqrt{-18i}=3-3i\)
\(\displaystyle z=2-i, \;\ z=-1-i\)
There are two solutions, as you know.
\(\displaystyle z=\frac{-(1+3i)\pm\sqrt{(1+3i)^{2}-4(1-i)(-4+2i)}}{2(1-i)}\)
These whittle down to simple solutions. Nothing messy. \(\displaystyle \sqrt{-18i}=3-3i\)
\(\displaystyle z=2-i, \;\ z=-1-i\)
Re: Complex Numbers
Just a matter of algebra.
\(\displaystyle \sqrt{-18i}=\sqrt{18}\sqrt{-i}\)
\(\displaystyle \sqrt{18}=3\sqrt{2}\)
\(\displaystyle \sqrt{-i}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i\)
\(\displaystyle \left(3\sqrt{2}\right)\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i\right)=3-3i\)
Just a matter of algebra.
\(\displaystyle \sqrt{-18i}=\sqrt{18}\sqrt{-i}\)
\(\displaystyle \sqrt{18}=3\sqrt{2}\)
\(\displaystyle \sqrt{-i}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i\)
\(\displaystyle \left(3\sqrt{2}\right)\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i\right)=3-3i\)