## frx 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? one year ago one year ago

1. Yahoo!

u can use quad formula here

2. frx

What happens then when i take the square root of -7/4+6i?

3. frx

$\sqrt{-\frac{ 7 }{4 }+6i}$

4. Yahoo!

$Z = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a }$

5. Yahoo!

$i^2 = -1$

6. frx

$\frac{ \-1+6i \pm \sqrt{(1-6i)^{2}+4(7*9i)} }{ 4}$

7. frx

But 1-6i squared is 1-12i-36 so i still get the root of i

8. hartnn

can't u use calculator ?

9. frx

We aren't allowed to use calculators at all at my university

10. hartnn

then its not easy to solve this: even with completing the square method, u need square root of complex number.

11. hartnn

can u convert complex number to polar form manually ?

12. frx

Yepp, maybe it's easier to solve in Euler form?

13. hartnn

not tried, ever

14. frx

Ok, I'll try it out it might just work :)

15. hartnn

u know the answer ? if not wanna know ?

16. frx

I know it, z=1+5i and z=-2+i

17. frx

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..

18. hartnn

yeah, even i did see wolf, all 3 methods....each required sqrt of (-7/4+6i)

19. frx

By completing the squares it can also be written as$(z+\frac{ 1-6i }{ 2 })^{2}-(\frac{ 1-6i }{2 })^{2} - (7+9i) = 0$

20. 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

21. frx

Ohh, that's clever! Will try it out!

22. phi

trust me, I did not figure this idea out first.

23. frx

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?

24. 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)

25. frx

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 ;)

26. frx

By the way are you guys at MIT allowed to use calculators to there kind of problems?

27. phi

I graduated a while ago. But in general I would guess yes. Lots of times tests were open book (not that it helped!)

28. frx

Open Book seems at least more genuine, it reflects the worklife better, than our memorization tests.

29. phi

your discriminant simplifies to -7+24i the square root is reasonably straight forward