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Solving Equations of Complex Numbers

  • #1

Homework Statement


Show that (1+i) is a root of the equation z4=-4 and find the other roots in the form a+bi where (a) and (b) are real.


Homework Equations


Using De Moivre's Theorem
zn=[rn,nθ]
Modulus(absolute value of z) = 4
Argument = ???

The Attempt at a Solution



r4=4 → r = (4)^(1/5)
argument(z) = ∏/4 (not sure if that was right....)

Let:
[r4],5θ] = [4,2n∏ + ∏/4]

and then solve for the solutions for (n=-1,-2,1,2...im assuming to make it symmetrical)
 

Answers and Replies

  • #2
1,796
53

Homework Statement


Show that (1+i) is a root of the equation z4=-4 and find the other roots in the form a+bi where (a) and (b) are real.


Homework Equations


Using De Moivre's Theorem
zn=[rn,nθ]
Modulus(absolute value of z) = 4
Argument = ???

The Attempt at a Solution



r4=4 → r = (4)^(1/5)
argument(z) = ∏/4 (not sure if that was right....)

Let:
[r4],5θ] = [4,2n∏ + ∏/4]

and then solve for the solutions for (n=-1,-2,1,2...im assuming to make it symmetrical)
No.

Where is 1+i in the complex plane? You can draw it, then the absolute value is just [itex]\sqrt{2}[/itex] and the principle argument, [itex]\theta[/itex], is [itex]\pi/4[/itex] right? So then [itex]1+i=\sqrt{2} e^{\pi i/4}[/itex]. Now what do you get when you raise that expression to the fourth power?

And in general, when you have an expression [itex] z^n=-k[/itex], we obtain [itex]z=(-k)^{1/n}[/itex] and

[tex](-k)^{1/n}=k^{1/n}e^{i/n(\pi+2j\pi)}, \quad j=0,1,2,\cdots,n-1[/tex]
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,833
955
Personally, I wouldn't bother with "polar form". The first part asks you to "Show that (1+i) is a root of the equation [itex]z^4=-4[/itex]". That, alone, does NOT ask you to solve the equation. You can "show that 'a' is a root of the equation f(x)= b" by evaluating f(a) and showing that it is 'b'. Here, if z= 1+ i then [itex]z^2= (1+ i)^2= 1^2+ 2(1)(i)+ i^2= 1+ 2i- 1= 2i[/itex]. Then [itex]z^4= ((1+ i)^2)^2= (2i)^2= -4[/itex].

You should also know that "if a+ bi is a root of a polynomial equation with real coefficients then the complex conjugate, a- bi is also a root". Having determined at 1+i is a root, we immediately know that 1- i is also a root. Knowing that 1+ i is a root, we know that z- (1+ i) is a factor of the polynomial [itex]z^4+ 4[/itex] and knowing that 1- i is a root, we know that z- (1- i) is a factor. That is, we can write [itex]z^4+ 4= (z- (1+i))(z- (1- i))P(x)[/itex].

But [itex](z- (1+i))(z- (1- i))= ((z- 1)- i)((z- 1)+ i)= (z- 1)^2- i^2= z^2- 2z+ 1- i^2= z^2- 2z+ 2[/itex] so that [itex]z^4+ 4= (z^2- 2z+ 2)P(z)[/itex]. Dividing [itex]z^4+ 4[/itex] by [itex]z^2- 2z+ 2[/itex] gives [itex]P(z)= z^2+ 2z+ 2[/itex]. Use the quadratic formula to solve [itex]P(z)= z^2+ 2z+ 2= 0[/itex].
 
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