anonymous
  • anonymous
Solve the complex equation: \[(2+i)z ^{2}+(8-11i)z-5-25i=0\] I started to isolate the first term and got: \[z ^{2}+\frac{ 8-11i }{ 2+i }z + \frac{ -5-25i }{ 2+1 } = 0\] \[z ^{2}+(1-6i)z-(7+9i)=0\] How should i continue from here? I can't use the quadratic formula, right? So i guess i need to factorize, but how?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
u can use quad formula here
anonymous
  • anonymous
What happens then when i take the square root of -7/4+6i?
anonymous
  • anonymous
\[\sqrt{-\frac{ 7 }{4 }+6i}\]

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anonymous
  • anonymous
\[Z = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a }\]
anonymous
  • anonymous
\[i^2 = -1\]
anonymous
  • anonymous
\[\frac{ \-1+6i \pm \sqrt{(1-6i)^{2}+4(7*9i)} }{ 4}\]
anonymous
  • anonymous
But 1-6i squared is 1-12i-36 so i still get the root of i
hartnn
  • hartnn
can't u use calculator ?
anonymous
  • anonymous
We aren't allowed to use calculators at all at my university
hartnn
  • hartnn
then its not easy to solve this: even with completing the square method, u need square root of complex number.
hartnn
  • hartnn
can u convert complex number to polar form manually ?
anonymous
  • anonymous
Yepp, maybe it's easier to solve in Euler form?
hartnn
  • hartnn
not tried, ever
anonymous
  • anonymous
Ok, I'll try it out it might just work :)
hartnn
  • hartnn
u know the answer ? if not wanna know ?
anonymous
  • anonymous
I know it, z=1+5i and z=-2+i
anonymous
  • anonymous
It got pretty wierd using the Euler formula so it's probobly not useble. But the Wolframalpha solutions factorize the hole thing into \[(z-1-5i)(z+(2-i))=0\] But I don't see how to come up with that without any computer..
hartnn
  • hartnn
yeah, even i did see wolf, all 3 methods....each required sqrt of (-7/4+6i)
anonymous
  • anonymous
By completing the squares it can also be written as\[(z+\frac{ 1-6i }{ 2 })^{2}-(\frac{ 1-6i }{2 })^{2} - (7+9i) = 0\]
phi
  • phi
one way to find the square root of the discriminant is use the idea (a+bi)^2 = sqrt( -117 + 44i) (Here I am using a, b, and c from the original problem) this means a^2-b^2 +2abi = -117 + 44i you can solve for a and b using a^2-b^2 = real part = -117 2ab = imag part = 44
anonymous
  • anonymous
Ohh, that's clever! Will try it out!
phi
  • phi
trust me, I did not figure this idea out first.
anonymous
  • anonymous
I've seen something like it before in my book so I get the idea :) But is this the general way to solve these kind of problems?
phi
  • phi
that idea is the basis for formulas for finding the root of a complex number in rect coords you could also change to polar. But that really needs a calculator to do the atans And in general, you are not going to get nice numbers no matter what approach (i.e. numbers you can handle without a calculator). This problem was chosen to be doable by hand (with a lot of work)
anonymous
  • anonymous
I really don't get why we are not allowed to use calculators but i guess it's good for the brain to do there problems without it. Is quite hard to convert into polarform to since the argument gets really nasty. Anyway, thank you alot for your help will have to practice this for som time i guess ;)
anonymous
  • anonymous
By the way are you guys at MIT allowed to use calculators to there kind of problems?
phi
  • phi
I graduated a while ago. But in general I would guess yes. Lots of times tests were open book (not that it helped!)
anonymous
  • anonymous
Open Book seems at least more genuine, it reflects the worklife better, than our memorization tests.
phi
  • phi
your discriminant simplifies to -7+24i the square root is reasonably straight forward

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