Calculating the roots of a quadratic with complex coefficien

1. Oct 24, 2015

astrololo

1. The problem statement, all variables and given/known data
Calculating the roots of a quadratic with complex coefficients

2. Relevant equations
x^2 - (5i+14)x+2(5i+12)=0

3. The attempt at a solution

I tried the quadratic solution but it gives too complicated solutions. I have no idea on how to do this...

2. Oct 24, 2015

Student100

You have something in the form of $$z^2=c$$ where $$c=a+bi$$ assume that a square root of c is y+zi, where $$(y+zi)^2=a+bi$$ Do you understand where to go from here? There are multiple methods to solve these problems, some easier than others.

Have you learned DeMoivre’s Theorem yet?

3. Oct 24, 2015

astrololo

No, I didn't learn it.

4. Oct 24, 2015

astrololo

The c^2=a+bi corresponds to what I had earlier, right ?

5. Oct 24, 2015

Student100

Okay.

Is the first part at least recognizable, do you know how to develop it into a formula for y and z?

6. Oct 24, 2015

Student100

No you just had c, you're looking for $$\sqrt(c)$$

7. Oct 24, 2015

astrololo

Sorry, I was referring to this : z^2=18.75+25i

ISn't this good ?

where z=x-2.5i-7

8. Oct 24, 2015

Student100

Looks okay.

You're looking for a $$(y+zi)^2=18.75+25i$$ equate the real and imaginary parts.

Last edited: Oct 25, 2015
9. Oct 24, 2015

astrololo

I don't understand. You want me to do this : y=18.75 and zi=25i ???

10. Oct 24, 2015

astrololo

bump

11. Oct 24, 2015

Student100

I just made it up, it's a dummy variable.

No, hang in there with me for a second, I want you to expand $(y+zi)^2$ first. Then equate the parts that still have an i to the imaginary part, and the parts of the equation that have no i to the real part as two equations.

12. Oct 24, 2015

astrololo

Ok, one step at a time : http://www4f.wolframalpha.com/Calculate/MSP/MSP64521di37fi0318f781f00005eih08d4i137gf4g?MSPStoreType=image/gif&s=22&w=132.&h=18. [Broken]

Which gives y^2+2yzi-z^2

Last edited by a moderator: May 7, 2017
13. Oct 24, 2015

Student100

Excellent. Now combine like terms and tell me which are the real and complex parts and equate them to the right side respectfully.

Last edited by a moderator: May 7, 2017
14. Oct 24, 2015

astrololo

y^2+2yzi-z^2

y^2-z^2 is the real part and 2yzi is the complex part.

15. Oct 24, 2015

Student100

So the real part is equal to your first equation, and the complex part is equal to second, why don't you go ahead and write that out.

16. Oct 24, 2015

astrololo

y^2-z^2=18,75

2yzi=25i

17. Oct 25, 2015

Student100

Yes, so a=18.75, and b =25.

So now you have two questions, solving equation 2 for $$z=\frac{b}{2y}$$ subbing it into equation one gives you $$y^2-(\frac{b}{2y})^2=a$$ which can be rewritten as $$4y^4-4ay^2-b^2=0$$ taking the positive root $$y^2=\frac{a+\sqrt{a^2+b^2}}{2}$$ which gives you $$y=\frac{1}{\sqrt{2}}\sqrt{a+\sqrt{a^2+b^2}}$$

That's the part of your real root, see if you can develop the complex part.

18. Oct 25, 2015

PeroK

You shouldn't have given up on the quadratic formula so easily. With complex numbers, things tend to get algebraically messy, so you have to accept more complications than with real numbers. In any case, using the quadratic formula directly gives:

$x = \frac{14 + 5i \pm 5 \sqrt{3 + 4i}}{2}$

Which is really not very complicated at all!