Calculating the roots of a quadratic with complex coefficien

[astrololo]

Homework Statement

Calculating the roots of a quadratic with complex coefficients

Homework Equations

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

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...

 

[Student100]

Homework Statement

Calculating the roots of a quadratic with complex coefficients

Homework Equations

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

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...

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?

 

[astrololo]

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?

No, I didn't learn it.

 

[astrololo]

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?

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

 

[Student100]

No, I didn't learn it.

Okay.

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

 

[Student100]

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

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

 

[astrololo]

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

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

ISn't this good ?

where z=x-2.5i-7

 

[Student100]

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

ISn't this good ?

where z=x-2.5i-7

Looks okay.

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

 

[astrololo]

First part looks okay, the second part isn't correct that's not the square root,

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

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

 

[astrololo]

bump

 

[Student100]

edit

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

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

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.

 

[astrololo]

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

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

 

[Student100]

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

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

 

[astrololo]

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

y^2+2yzi-z^2

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

 

[Student100]

y^2+2yzi-z^2

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

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.

 

[astrololo]

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.

y^2-z^2=18,75

2yzi=25i

 

[Student100]

y^2-z^2=18,75

2yzi=25i

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.

 

[PeroK]

Homework Statement

Calculating the roots of a quadratic with complex coefficients

Homework Equations

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

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...

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!