Homework Help: Solving complex number equations

1. Sep 26, 2005

deryk

Hello, Im trying to solve

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

(z^2+1-i)=0

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

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: Sep 26, 2005
2. Sep 26, 2005

LeonhardEuler

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. Sep 26, 2005

HallsofIvy

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):
$$z= \sqrt{-1+ i}$$

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 re&theta;i where
$r= sqrt{a^2+ b^2}$ and $\theta= tan^{-1}(\frac{b}{a})$.

4. Sep 27, 2005

deryk

How would you write z=+ or -(-1+i)^0.5 & z= + or - (2-i)^0.5 in polar and cartesian form? Thanks.

5. Sep 27, 2005

hypermorphism

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. Sep 27, 2005

HallsofIvy

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
$$r^{\frac{1}{2}}(cos\frac{\theta}{2}+ i sin\frac{\theta}{2})$$
and
$$r^{\rac{1}{2}}(cos(\frac{\theta}{2}+ \pi)+ i sin(\frac{\theta}{2}+\pi))$$
or
$$r^{\frac{1}{2}}e^{\frac{\theta}{2}}$$
and
$$r^{\frac{1}{2}}e^{(\frac{\theta}{2}+\pi)}$$