6. Példa $\sqrt{-15+8i}=?$

Megoldás.

a) $\sqrt{-15+8i}=z=a+bi \Rightarrow 15-(a^2-b^2)+(-8+2ab)i=0\Rightarrow$

$a^2-b^2=15$

$2ab=8$

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$b={4\over a}\Rightarrow a^2-b^2=a^2-{16\over{a^2}}=15\Rightarrow$

$(a^2)^2-15(a^2)-16=0 \Rightarrow (a^2)_1={15+17\over2},(a^2)_2={15-17\over2}(<0)\Rightarrow a^2=16$

$a_1=4, b_1=1$

$a_2=-4, b_2=-1\Rightarrow$

$z_1=1+4i, z_2=-1-4i$

b) $-15+8i=17(-{15\over{17}}+{8\over{17}}i)\approx17(\cos(2.652)+i\sin(2.652))\Rightarrow$

$z_1=\sqrt{17}(\cos({2.652+0\over2})+i\sin({2.652+0\over2}))\approx1+4i$

$z_2=\sqrt{17}(\cos({2.652+2\pi)\over2})+i\sin({2.652+2\pi)\over2}))\approx-1-4i$