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Solving complex number equations

  • Thread starter deryk
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  • #1
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Hello, Im trying to solve

(z^4-2+i)(z^2+1-i)=0

With the quadratic formula:

(z^2+1-i)=0

Does a=1, b=1 & c=-1?

Thanks for your time.

IM meant to

(a) Give answers in polar form using the principal argument;
(b) Give answers in cartesian form

Cartesian is (x,y) is it not and polar is (r,theta) Im just not sure about the principal argument.
 
Last edited:

Answers and Replies

  • #2
LeonhardEuler
Gold Member
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deryk said:
Does a=1, b=1 & c=-1?

Thanks for your time.
No, a=1, b=0, and c=(1-i) because "a" is the coefficient of z^2, "b" is the coefficient of z and "c" is the constant term.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
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deryk said:
Hello, Im trying to solve

(z^4-2+i)(z^2+1-i)=0

With the quadratic formula:

(z^2+1-i)=0

Does a=1, b=1 & c=-1?

Thanks for your time.

IM meant to

(a) Give answers in polar form using the principal argument;
(b) Give answers in cartesian form

Cartesian is (x,y) is it not and polar is (r,theta) Im just not sure about the principal argument.
To answer your questin a= 1, b= 1, and c= -i.
However, there is no reason to use the quadratic formula here- since
z2+ 1-i= 0, z2= -1+ i and you just take the square root (this is what the quadratic formula would give you anyway):
[tex]z= \sqrt{-1+ i}[/tex]

You'll want to convert that into polar form to take the square root. The answer will then be in polar form and you will need to convert back to Cartesian form.
The polar form of a+ bi is reθi where
[itex]r= sqrt{a^2+ b^2}[/itex] and [itex]\theta= tan^{-1}(\frac{b}{a})[/itex].
 
  • #4
10
0
HallsofIvy said:
To answer your questin a= 1, b= 1, and c= -i.
However, there is no reason to use the quadratic formula here- since
z2+ 1-i= 0, z2= -1+ i and you just take the square root (this is what the quadratic formula would give you anyway):
[tex]z= \sqrt{-1+ i}[/tex]

You'll want to convert that into polar form to take the square root. The answer will then be in polar form and you will need to convert back to Cartesian form.
The polar form of a+ bi is reθi where
[itex]r= sqrt{a^2+ b^2}[/itex] and [itex]\theta= tan^{-1}(\frac{b}{a})[/itex].
How would you write z=+ or -(-1+i)^0.5 & z= + or - (2-i)^0.5 in polar and cartesian form? Thanks.
 
  • #5
deryk said:
How would you write z=+ or -(-1+i)^0.5 & z= + or - (2-i)^0.5 in polar and cartesian form? Thanks.
Pretend the complex number under the radicand is a 2-dimensional position vector with components (-1,1) in the case of the first one. Get the argument (angle from the positive real axis) and modulus (magnitude of the vector) from that, then write it in polar form to get the roots.
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,770
911
deryk said:
How would you write z=+ or -(-1+i)^0.5 & z= + or - (2-i)^0.5 in polar and cartesian form? Thanks.
Since you copied the formulas I gave for converting from cartesian to polar coordinates, I assume you are asking about the square root:

DeMoivre's formula: If a complex number has "polar coordinates" r, θ (in other words, the number can be written r(cos θ+ i sin θ) or re) then it's square roots are
[tex]r^{\frac{1}{2}}(cos\frac{\theta}{2}+ i sin\frac{\theta}{2})[/tex]
and
[tex]r^{\rac{1}{2}}(cos(\frac{\theta}{2}+ \pi)+ i sin(\frac{\theta}{2}+\pi))[/tex]
or
[tex]r^{\frac{1}{2}}e^{\frac{\theta}{2}}[/tex]
and
[tex]r^{\frac{1}{2}}e^{(\frac{\theta}{2}+\pi)}[/tex]
 
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