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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?
 one year ago
 one year ago
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?
 one year ago
 one year ago

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Yahoo!Best ResponseYou've already chosen the best response.1
u can use quad formula here
 one year ago

frxBest ResponseYou've already chosen the best response.0
What happens then when i take the square root of 7/4+6i?
 one year ago

frxBest ResponseYou've already chosen the best response.0
\[\sqrt{\frac{ 7 }{4 }+6i}\]
 one year ago

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

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

frxBest ResponseYou've already chosen the best response.0
But 16i squared is 112i36 so i still get the root of i
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
can't u use calculator ?
 one year ago

frxBest ResponseYou've already chosen the best response.0
We aren't allowed to use calculators at all at my university
 one year ago

hartnnBest 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.
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
can u convert complex number to polar form manually ?
 one year ago

frxBest ResponseYou've already chosen the best response.0
Yepp, maybe it's easier to solve in Euler form?
 one year ago

frxBest ResponseYou've already chosen the best response.0
Ok, I'll try it out it might just work :)
 one year ago

hartnnBest ResponseYou've already chosen the best response.0
u know the answer ? if not wanna know ?
 one year ago

frxBest ResponseYou've already chosen the best response.0
I know it, z=1+5i and z=2+i
 one year ago

frxBest 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..
 one year ago

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

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

phiBest 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
 one year ago

frxBest ResponseYou've already chosen the best response.0
Ohh, that's clever! Will try it out!
 one year ago

phiBest ResponseYou've already chosen the best response.1
trust me, I did not figure this idea out first.
 one year ago

frxBest 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?
 one year ago

phiBest 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)
 one year ago

frxBest 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 ;)
 one year ago

frxBest 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?
 one year ago

phiBest 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!)
 one year ago

frxBest ResponseYou've already chosen the best response.0
Open Book seems at least more genuine, it reflects the worklife better, than our memorization tests.
 one year ago

phiBest ResponseYou've already chosen the best response.1
your discriminant simplifies to 7+24i the square root is reasonably straight forward
 one year ago
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