I'm asked to solve this equation : $$z^4+7(1+i)z^2+25i=0$$ by taking $z^2 = u $ : $$u^2+7(1+i)u+25i=0$$ this turns out to be a quadratic equation. solving it gives: $$u_1=-6-8i$$ $$u_2=-8-6i$$ so how are we supposed to solve $z^2=-6-8i$ and $z^2=-8-6i$ ? answeres are : $$z=\pm \sqrt5(sin\beta/2-icos\beta/2)$$ $$z=\pm \sqrt5(sin\alpha/2-icos\alpha/2)$$ where $\alpha = Arctan(4/3)$ and $\beta = Arctan(3/4)$ I have no idea how to get this answers.
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3Put $u_1$ and $u_2$ in polar form then use DeMoivre's Theorem for the roots – Paul Nov 27 '15 at 20:31
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Hint: $$ u^2+(7-7i)u+25i=(u+3+4i)(u+4+3i) $$ so you have some mistake in your solutions.
Write the solutions in polar form:
$$ u_1=-3-4i=5e^{i\theta_1} \quad with \quad \theta_1=\arctan \frac{4}{3} $$ $$ u_2=-4-3i=5e^{i\theta_1} \quad with \quad \theta_2=\arctan \frac{3}{4} $$ now you can do ?
Emilio Novati
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$$z^4+7(1+i)z^2+25i=0\Longleftrightarrow$$ $$z^4+(7+7i)z^2+25i=0\Longleftrightarrow$$
Substitute $x=z^2$:
$$x^2+(7+7i)x+25i=0\Longleftrightarrow$$ $$(x+(3+4i))(x+(4+3i))=0\Longleftrightarrow$$ $$x+(3+4i)=0\Longleftrightarrow\space\space\vee\space\space x+(4+3i)=0\Longleftrightarrow$$ $$x=-3-4i\Longleftrightarrow\space\space\vee\space\space x=-4-3i\Longleftrightarrow$$ $$z^2=-3-4i\Longleftrightarrow\space\space\vee\space\space z^2=-4-3i\Longleftrightarrow$$ $$z^2=|-3-4i|e^{\arg(-3-4i)i}\Longleftrightarrow\space\space\vee\space\space z^2=|-4-3i|e^{\arg(-4-3i)i}\Longleftrightarrow$$ $$z^2=5e^{\left(\arctan\left(\frac{4}{3}\right)-\pi\right)i}\Longleftrightarrow\space\space\vee\space\space z^2=5e^{\left(\arctan\left(\frac{3}{4}\right)-\pi\right)i}\Longleftrightarrow$$ $$z=\left(5e^{\left(\arctan\left(\frac{4}{3}\right)-\pi+2\pi k\right)i}\right)^{\frac{1}{2}}\Longleftrightarrow\space\space\vee\space\space z=\left(5e^{\left(\arctan\left(\frac{3}{4}\right)-\pi+2\pi l\right)i}\right)^{\frac{1}{2}}\Longleftrightarrow$$ $$z=\sqrt{5}e^{\frac{1}{2}\left(\arctan\left(\frac{4}{3}\right)-\pi+2\pi k\right)i}\space\space\vee\space\space z=\sqrt{5}e^{\frac{1}{2}\left(\arctan\left(\frac{3}{4}\right)-\pi+2\pi l\right)i}$$
With $k,l\in\mathbb{Z}$ and $k,l:0-1$
So the solutions are:
$$z_0=\sqrt{5}e^{\frac{1}{2}\left(\arctan\left(\frac{4}{3}\right)-\pi+2\pi \cdot 0\right)i}=1-2i$$ $$z_1=\sqrt{5}e^{\frac{1}{2}\left(\arctan\left(\frac{4}{3}\right)-\pi+2\pi \cdot 1\right)i}=-1+2i$$ $$z_2=\sqrt{5}e^{\frac{1}{2}\left(\arctan\left(\frac{3}{4}\right)-\pi+2\pi \cdot 0\right)i}=\sqrt{-4-3i}$$ $$z_3=\sqrt{5}e^{\frac{1}{2}\left(\arctan\left(\frac{3}{4}\right)-\pi+2\pi \cdot 1\right)i}=-\sqrt{-4-3i}$$
Jan Eerland
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