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Homework Help
Calculus and Beyond Homework Help
Solving Complex Equations: z2+2(1-i)z+7i=0
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[QUOTE="KUphysstudent, post: 6066241, member: 573120"] [h2]Homework Statement [/h2] So it is pretty straight forward, solve this. z[SUP]2[/SUP]+2(1-i)z+7i=0 [h2]Homework Equations[/h2] z[SUP]2[/SUP]+2(1-i)z+7i=0 (-b±√(b[SUP]2[/SUP]-4ac))/2a [h2]The Attempt at a Solution[/h2] So what I would do first is solve 2(2-1)z, I get (2-2i)z=2z-2iz we now have z[SUP]2[/SUP]-2iz+7i+2z=0 Now I don't really know what to do because my textbook has two examples, in both z[SUP]2[/SUP] is ignored. first it has z[SUP]2[/SUP]+2iz-1-i=0 and used a=1, b=2i and c=-1-i the second example shows z[SUP]2[/SUP]+2z+4=0 and has 2z=b and ac =4*1 the problem with the two examples is I cannot deduce what will be a, b, and c in my problem. I mean following the logic of the first example I get 2z = b or -2iz = b and then c = either 7i or 7i + 2z or something completely different. I tried plugging in the numbers so: (-2i ±√(2i^2-4*7i))/2 = (-2i±√(-4-28i))/2 then I tried 2z = b instead of 2i = b (-2 ±√(2^2-4*7i))/2 = (-2±√(4-28i))/2 I mean this is just guessing and I kept going, what if 7i = b, etc. but it doesn't help me understand what it should be and why, which is really what I want to know and not the solution. [/QUOTE]
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Solving Complex Equations: z2+2(1-i)z+7i=0
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