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frx
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Solve the complex equation: \[(2+i)z ^{2}+(811i)z525i=0\]
I started to isolate the first term and got:
\[z ^{2}+\frac{ 811i }{ 2+i }z + \frac{ 525i }{ 2+1 } = 0\]
\[z ^{2}+(16i)z(7+9i)=0\]
How should i continue from here? I can't use the quadratic formula, right? So i guess i need to factorize, but how?
 2 years ago
 2 years ago
frx Group Title
Solve the complex equation: \[(2+i)z ^{2}+(811i)z525i=0\] I started to isolate the first term and got: \[z ^{2}+\frac{ 811i }{ 2+i }z + \frac{ 525i }{ 2+1 } = 0\] \[z ^{2}+(16i)z(7+9i)=0\] How should i continue from here? I can't use the quadratic formula, right? So i guess i need to factorize, but how?
 2 years ago
 2 years ago

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Yahoo! Group TitleBest ResponseYou've already chosen the best response.1
u can use quad formula here
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.0
What happens then when i take the square root of 7/4+6i?
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.0
\[\sqrt{\frac{ 7 }{4 }+6i}\]
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.1
\[Z = \frac{ b \pm \sqrt{b^2  4ac} }{ 2a }\]
 2 years ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.1
\[i^2 = 1\]
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ \1+6i \pm \sqrt{(16i)^{2}+4(7*9i)} }{ 4}\]
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.0
But 16i squared is 112i36 so i still get the root of i
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
can't u use calculator ?
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.0
We aren't allowed to use calculators at all at my university
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
then its not easy to solve this: even with completing the square method, u need square root of complex number.
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
can u convert complex number to polar form manually ?
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Yepp, maybe it's easier to solve in Euler form?
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
not tried, ever
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Ok, I'll try it out it might just work :)
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
u know the answer ? if not wanna know ?
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.0
I know it, z=1+5i and z=2+i
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.0
It got pretty wierd using the Euler formula so it's probobly not useble. But the Wolframalpha solutions factorize the hole thing into \[(z15i)(z+(2i))=0\] But I don't see how to come up with that without any computer..
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
yeah, even i did see wolf, all 3 methods....each required sqrt of (7/4+6i)
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.0
By completing the squares it can also be written as\[(z+\frac{ 16i }{ 2 })^{2}(\frac{ 16i }{2 })^{2}  (7+9i) = 0\]
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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^2b^2 +2abi = 117 + 44i you can solve for a and b using a^2b^2 = real part = 117 2ab = imag part = 44
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Ohh, that's clever! Will try it out!
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
trust me, I did not figure this idea out first.
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.0
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?
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
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)
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.0
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 ;)
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.0
By the way are you guys at MIT allowed to use calculators to there kind of problems?
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
I graduated a while ago. But in general I would guess yes. Lots of times tests were open book (not that it helped!)
 2 years ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Open Book seems at least more genuine, it reflects the worklife better, than our memorization tests.
 2 years ago

phi Group TitleBest ResponseYou've already chosen the best response.1
your discriminant simplifies to 7+24i the square root is reasonably straight forward
 2 years ago
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