קיץ 2023 · מועד מיוחד

$z^2 = (x+iy)^2 = x^2 + 2xyi + i^2y^2$

$i^2 = -1 \;\Rightarrow\; z^2 = (x^2 - y^2) + 2xy\,i$

$z^2 - 3i = (x^2-y^2) + (2xy-3)i$

$z^2 + 5i = (x^2-y^2) + (2xy+5)i$

$|z^2-3i| = \sqrt{(x^2-y^2)^2 + (2xy-3)^2}$

$|z^2+5i| = \sqrt{(x^2-y^2)^2 + (2xy+5)^2}$

$(2xy-3)^2 = (2xy+5)^2$

$4x^2y^2 - 12xy + 9 = 4x^2y^2 + 20xy + 25$

$-12xy + 9 = 20xy + 25$

$-32xy = 16$

$xy = -\dfrac{1}{2}$

$y = -\dfrac{1}{2x}$

$y = -\dfrac{1}{2\cdot 1} = -\dfrac{1}{2}$

$z = 1 - \dfrac{1}{2}i$

$1 = \text{cis}(0° + 360°k)$

$z = \sqrt[6]{\text{cis}(0°+360°k)} = \text{cis}\left(\dfrac{0°+360°k}{6}\right)$

$z_k = \text{cis}(60°k), \;\; k = 0,1,2,3,4,5$

$z_1 = \text{cis}0°, \quad z_2 = \text{cis}60°, \quad z_3 = \text{cis}120°$

$z_4 = \text{cis}180°, \quad z_5 = \text{cis}240°, \quad z_6 = \text{cis}300°$

$4x^4 - 4x^2 + 1 = 0$

$(2x^2 - 1)^2 = 0$

$2x^2 - 1 = 0 \;\Rightarrow\; x^2 = \dfrac{1}{2}$

$x = \pm\dfrac{\sqrt2}{2}$

$y_A = -\dfrac{1}{2\cdot\frac{\sqrt2}{2}} = -\dfrac{1}{\sqrt2} = -\dfrac{\sqrt2}{2}$

$A = \dfrac{\sqrt2}{2} - \dfrac{\sqrt2}{2}i$

$r = |A| = \sqrt{\left(\tfrac{\sqrt2}{2}\right)^2 + \left(\tfrac{\sqrt2}{2}\right)^2} = \sqrt{\tfrac12 + \tfrac12} = 1$

$A = \text{cis}(-45°)$

$\text{cis}15°, \quad \text{cis}75°, \quad \text{cis}135°$

$\text{cis}195°, \quad \text{cis}255°, \quad \text{cis}315°$

$60°k + \alpha = 15° + 60°k \;\Rightarrow\; \alpha = 15°$