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How do I solve this equation? Also plotting all these solutions on the graph needed.

$z^2+\bar z^2=0$

I have got something like

$(x+iy)^2=\sqrt{-(x-iy)^2}$

$x+iy=y+ix$

What can do next? Help Please!

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    $\begingroup$ Try the polar coordinates $z=re^{\phi i}$. $\endgroup$ Aug 2 '17 at 11:11
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Considering that $\overline{z}^2=\overline{z^2}$, then $$ 0=z^2+\overline{z^2}=2 \mathrm{Re}(z^2) \Longleftrightarrow \mathrm{Re}(z^2)=0 \Longleftrightarrow |\mathrm{Re}(z)|=|\mathrm{Im}(z)|. $$

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  • $\begingroup$ How can I plot all these solutions on the graph? $\endgroup$
    – markable
    Aug 2 '17 at 20:46
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    $\begingroup$ If $z=x+iy$ then $|x|=|y|$ means the in the $xy$-plane you are plotting the two diagonals $\endgroup$ Aug 2 '17 at 20:50
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Let $z \in \mathbb{C}$.

$ \begin{array}{rcl} z^2 + \bar z^2 = 0 & \Leftrightarrow & (z+\bar z)^2 = 2z \bar z \\ & \Leftrightarrow & (2 \mathcal{R}e(z))^2 = 2 |z|^2 \\ & \Leftrightarrow & 2 \mathcal{R}e(z)^2 = \mathcal{R}e(z)^2 + \mathcal{I}m(z)^2 \\ & \Leftrightarrow & \mathcal{R}e(z)^2 = \mathcal{I}m(z)^2 \end{array} $

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Let $z=x+iy $ substituting we have $x^2+2ixy-y^2+x^2-2ixy-y^2=0$ thus $x^2=y^2$ hence $x=\pm y $ thus given equation represents a pair of straight lines namely $x=y,x=-y $

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You can also use the polar coordinates $z=re^{\phi i}$. Because $\bar{z}=re^{-\phi i}$ we get $$ 0=z^2+\bar{z}^2=r^2e^{2\pi i}+r^2e^{-2\phi i}=2r^2\cos(2\phi). $$ Therefore either $r=0$ or $\phi\in \frac{\pi}4+\frac{\pi}2\mathbb Z$. We conclude $$ z\in\{0, re^{\frac14\pi i}, re^{\frac34\pi i}, re^{\frac54\pi i}, re^{\frac74\pi i}~:~r>0\}. $$ This set contains the center $0$ and the four diagonal rays.

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$$0 = z^2+\bar{z}^2 = z^2 - (i\bar{z})^2 = (z + i\bar{z})(z - i\bar{z}).$$ Now, $$0 = z + i\bar{z} = x + yi + xi + y = (x + y)(1 + i)\iff x + y = 0,$$ and by a similar reasoning, $z - i\bar{z}\iff x = y$.

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What about, in polar, $z=re^{i\theta}$. So, $z^2=r^2e^{2i\theta}$, $\bar z^2=r^2e^{-2i\theta} $. So, $$z^2+\bar z^2=r^2 (e^{2i\theta}+e^{-2i\theta})=2r^2 \cos (2\theta)$$. So, $2r^2\cos (2\theta)=O$. Therefore $r=O$ or $\theta=(2k+1)\frac\pi4$. Thus we have lines through the origin at angles odd multiples of $\frac\pi4$. So, two lines : $re^{i\frac\pi4}$ and $re^{i\frac{3\pi}4} $ for $r\in (-\infty,\infty) $.

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