Complex numbered equation

Question

I came across this equation, I need to find all roots of it:

$$z^{8}+\left( \dfrac {\sqrt {2}}{2}-\dfrac {\sqrt {2}}{2}i\right) z^{4}+1-i=0$$

I substituted: $$z^{4}=w$$

The discriminant:

$$D=(\frac{\sqrt2}{2}-\frac{\sqrt2}{2})^2-4(1-i)=\cdots=-4+3i$$

When I want to solve the roots of the primary equation I cant solve: $\sqrt{-4+3i}$

If I put it in polar form the angle is $\phi=\arctan{(-3\pi/4)}$, a messy solution of arctangens.

Is my technique of tackling this problem wrong?

Answer 1

Let $$\sqrt{-4+3i}=a+ib$$ where $a,b$ are real

$-4+3i=a^2-b^2+2abi$

Equate the imaginary and the real parts, $2ab=3\implies a,b$ will have the same sign

$a^2-b^2=-4\implies-4=(3/2b)^2-b^2\iff4b^4-16b^2-9=0$

$b^2=\dfrac{16\pm\sqrt{16^2+16\cdot9}}8=\dfrac{16\pm20}8=\dfrac92$ as $b^2\ge0$

$\implies b=\pm\dfrac3{\sqrt2}$

$a=\dfrac3{2b}=?$

Answer 2

\begin{gather*} x^{8} +\left(\frac{\sqrt{2}}{2} -\frac{\sqrt{2}}{2} i\right) x^{4} +1-i=0\\ Substitute\ x^{4} \ =\ z\\ z^{2} +\left(\frac{\sqrt{2}}{2} -\frac{\sqrt{2}}{2} i\right) z+1-i=0\\ Now\ ,\ let\ x+iy\ =\ \sqrt{-4+3i}\\ x^{2} -y^{2} +2xyi\ =\ -4+3i\\ x^{2} -y^{2} \ =\ -4\ and\ 2xy\ =\ 3\\ x^{2} +y^{2} \ =\sqrt{\left( x^{2} -y^{2}\right)^{2} +( 2xy)^{2}} =\sqrt{4^{2} +3^{2}} =5\\ 2x^{2} \ =\ 1,\ x=\frac{1}{\sqrt{2}}\\ 2y^{2} \ =\ 9,\ y=\frac{3}{\sqrt{2}}\\ x+iy\ =\ \frac{1}{\sqrt{2}}( 1+3i)\\ Now,\ as\ per\ original\ question,\ \\ z=\frac{\frac{- i+1}{\sqrt{2}} +\frac{1+3i}{\sqrt{2}}}{2} =\frac{2+2i}{2\sqrt{2}} =\frac{1}{\sqrt{2}}( 1+i)\\ z\ =cos\frac{\pi }{4} +isin\frac{\pi }{4}\\ x^{4} \ =\ e^{i\pi /4}\\ x^{4} \ =\ e^{i( 2k\pi +\pi /4)}\\ x=e^{i( k\pi /2+\pi /16)} \end{gather*}}

Answer 3

It could be solved in polar form, in which the equation reads,

$$z^8 + e^{-i\frac\pi4} z^4+ \sqrt2 e^{-i\frac\pi4}= 0$$

Factorize,

$$(z^4 - \sqrt2 e^{i\frac\pi2} )(z^4 - e^{i\frac{5\pi}4})=0$$

which leads to $z^4 = \sqrt2 e^{i\frac\pi2}$ and $z^4 = e^{i\frac{5\pi}4}$. Thus, the solutions are

$$z = 2^{\frac18} e^{i(\frac{1+4n}8)\pi},\>\>\> e^{i(\frac{5+8n}{16})\pi} $$

Answer 4

The modulus of ${-4+3i}$ is $5.$ Treating the complex number as a position vector on the complex plane, we take a vector of equal length along the positive real axis, namely the real number $5,$ and take the sum with ${-4+3i}$. The result, ${1+3i}$. is a vertex of the rhombus whose other vertices are ${-4+3i}$, $0$, and $5.$ Hence it bisects the angle between ${-4+3i}$ and $5$ and therefore is parallel to $\sqrt{-4+3i}$.

We find that $(1+3i)^2=-8+6i$, which is exactly $2$ times the desired result of squaring, so we divide by $\sqrt2.$ And of course we can also multiply (or not) by $-1$. As a result, we find that $$ \sqrt{-4+3i} = \pm\frac{\sqrt2}{2} (1+3i),$$ which we can verify by directly computing the square.