What are the solutions to equations involving complex numbers?

In summary, the conversation discusses finding the real values of r that satisfy a given equation involving imaginary numbers, and determining the solutions for a similar question. The conversation includes a step-by-step guide and solutions for both equations.
  • #1
dcl
55
0
Heyas. I'm not too good at complex numbers so excuse me if these questions are a bit on the laughable side.

Find all real values for r for which ri is a solution of the equation.
z^4 - 2z^3 + 11z^2 - 18z + 18 = 0
hence, Determine all the solutions of the equations...

I'm not really sure how to tackle questions like this. I subbed in 'ri' for 'z' and worked out r, but it had multiple values, some of which are not solutiuons as listed in the back of the book. Perhaps i did it wrong all together.

A quick guide through the question would help heaps :)

Also

A similar question
Find the real number 'k' such that z=ki is a root of the equation:
z^3 + (2+i)z^2 + (2+2i)z + 4 = 0.

is this basically the same as the earlier question?


Thanks in advance :)
 
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  • #2
Okay, assuming you are given that there is an imaginary solution,
replace z with ri. Then, as I am sure you calculated, z^2= -r^2,
z^3= -r^3 i and z^4= r^4. Substituting those into the equation,
r^4+ 2r^3i- 11r^2- 18ri+ 18= 0.

Of course, in order for this to be 0, both real and imaginary parts must be 0. The real part is r^4- 11r^2+ 18= 0 and the imaginary part is 2r^3- 18r= 0. The crucial part is that the same r must satisfy BOTH equations. It's easy to factor the second equation:
2r^3- 18r= 2r(r^2- 9)= 0. The solutions are r=0, r= 3 and r= -3. It's obvious that r=0 does not satisfy r^4- 11r^2+ 18= 0. What about r= 3 and r= -3?

Now, for z^3 + (2+i)z^2 + (2+2i)z + 4 = 0.

Again, if z= ri, then z^2= -r^2 and z^3= -r^3 i. Substituting that into the equation, -r^3 i+ (2+i)(-r^2)+ (2+ 2i)(ri)+ 4=
-r^3 i- 2r^2- r^2 i+ 2ri- 2r+ 4= 0.

Separate real and imaginary parts: the real part is
-2r^2- 2r+ 4= 0. The imaginary part is -r^3- r^2+ 2r= 0.
The first of those factors as -2(r+2)(r-1)= 0 and so has solutions r= 1, r= -2. The second factors as -r(r^2+ r- 2)= -r(r+2)(r-1)= 0 and so has solutions r= 0, r= 1, r= -2. Remembering that r must satisfy BOTH equations, what can r be?
 
  • #3
Thanks for that man :)
 

1. What are complex numbers and how are they different from real numbers?

Complex numbers are numbers that contain both a real and imaginary component. They are written in the form a + bi, where a is the real part and bi is the imaginary part. Real numbers only contain a single numerical value and do not have an imaginary component.

2. How are complex numbers represented on a graph?

Complex numbers are represented on a graph called the complex plane. The real component is plotted on the horizontal axis and the imaginary component is plotted on the vertical axis.

3. How do you add, subtract, multiply, and divide complex numbers?

To add or subtract complex numbers, simply combine the real and imaginary components separately. To multiply, use the FOIL method and remember that i^2 = -1. To divide, multiply the numerator and denominator by the complex conjugate of the denominator and simplify.

4. What is the purpose of using complex numbers in problem-solving?

Complex numbers are often used in problem-solving to represent quantities that cannot be expressed using real numbers alone. They are also useful in solving equations and representing physical quantities such as electrical currents and wave functions.

5. Can complex numbers have a negative imaginary component?

Yes, complex numbers can have a negative imaginary component. The imaginary component is represented by the term bi, where b can be any real number. So, if b is negative, then the imaginary component will also be negative.

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