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Problem

When attempting to solve this question, I multiplied the equations to extrapolate the zw. Once I did this though, I ran into some confusion. Am I allowed to use the quadratic formula here? If I am, can someone demonstrate how to solve the part under the radical?

Dude156
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  • Yes you are. This is an example of why knowing the proof of something allows you to decide whether you can apply it in slightly different conditions. Note that any non-zero complex number has two square roots. – Arnaud Mortier Jun 01 '18 at 02:20
  • @ArnaudMortier I do know the quadratic formula proof -derived using completing the square version at least; I still don't understand how to clear the radical though – Dude156 Jun 01 '18 at 02:23

1 Answers1

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Yes you doing in the right way.

Given $z+\frac{20i}{w}=5+i$ and $w+\frac{12i}{z}=-4+10i$ $$zw+32i-\frac{240}{zw}=-30+46i$$ $$(zw)^2+(30-14i)zw-240=0$$

Now use the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ and we get

Here $a=1,b=(30-14i),c=-240$ $$=\frac{-(30-14i)\pm \sqrt{(30-14i)^2-4(1)(-240)}}{2(1)}$$ $$=\frac{-2(15-7i)\pm \sqrt{(2(15-7i))^2-4(1)(-240)}}{2(1)}$$ $$=\frac{-2(15-7i)\pm 2\sqrt{(15-7i)^2+240}}{2(1)}$$ Note that $\sqrt{(15-7i)^2+240}=\sqrt{416-210i}=\sqrt{(5i-21)^2}=5i-21$ $$7i-15\pm\sqrt{(15-7i)^2+240}=6+2i,12i-36$$

Now $$|zw|^2=6^2+2^2=40$$ and $$|zw|^2=12^2+36^2>40$$ So, the smallest value is $40$

tien lee
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