Show that the locus formed by $z$ in the equation $z^3+iz=1$ never crosses the coordinate axes in the Argand's plane.Further show that $|z|=\sqrt{\frac{-\text{Im}(z)}{2\text{Re}(z)\text{Im}(z)+1}}$    Let $z=x+iy$.Then $z^3+iz=1$ becomes $(x+iy)^3+i(x+iy)=1$ $x^3-iy^3+3ix^2y-3xy^2+ix-y=1$ $(x^3-3xy^2-y)+i(3x^2y-y^3+x)=1$ Comparing real and imaginary parts on both sides,we get $x^3-3xy^2-y=1,3x^2y-y^3+x=0$ But i do not know how to prove that the locus formed by $z$ never crosses the coordinate axes in the Argand's plane. In the second part,$|z|=\sqrt{x^2+y^2}$ and $\sqrt{\frac{-\text{Im}(z)}{2\text{Re}(z)\text{Im}(z)+1}}=\sqrt{\frac{-y}{2xy+1}}$.I don't understand how to prove them equal.
Show that the locus formed by z in the equation $z^3 + iz = 1$  never crosses the co-ordinate axes in the Argand plane. Further show that $$|z| = \sqrt{\dfrac{-\Im(z)}{2\Re(z)\Im(z) +1}}$$  Concept: $z=x+iy$ $z^3=(x+iy)^3=x^3-3xy^2+i(3x^2y-y^3)$ $iz=-y+ix$ Combined equation is $x^3-3xy^2-y-1+i(3x^2y-y^3+x)=0$ How do we proceed from here onward?
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What lens characteristics are important in star photography? 2 GROUP BY WITH DISTINCT , SUM , COUNT : PHP MYQL Ref :   I got answer (thanks coytech ) but i need one more column in its :  as per :  id | mid | pid | owgh | nwgh | 1    3      12    1.5    0.6 2    3      12    1.5    0.3 3    3      14    0.6    0.4 4    3      15    1.2    1.1 5    4      16    1.5    1.0 6    4      17    2.4    1.2  7    3      19    3.0    1.4   i got answer      Select mid ,          COUNT(distinct pid) as cpid  ,          SUM(nwgh) as totalnwgh  from test  GROUP BY mid   sqlfiddle :   mid  cpid       totalnwgh 3      4          3.8 4      2          2.2    But above i need one more column that's as below : totowgh  mid cpid        totalnwgh  totowgh 3      4          3.8        6.3 (DISTINCT value as per pid column) 4      2          2.2        3.9   where totowgh = 6.3 come by DISTINCT value of owgh column  as per distinct pid column  mid = 3 has count 5 but distinct pid  = 4 for mid=3 same way "distinct" owgh = 6.3 for mid=3 and distinct pid.  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Summation of natural number set with power of $m$ Who knows about the summation of this series: $$\sum\limits_{i=1}^{n}i^m $$  where $m$ is constant and $m\in \mathbb{N}$?  thanks So I integrate the holomorphic function $\frac{e^{iz} -1}{iz}$ over the half disk in the upper half plane, let me name it $\Gamma$. By using Cauchy's theorem, it equals 0.  $$0 = \int_{\Gamma} \frac{e^{iz} -1}{iz} = \int_{0}^{\pi}(e^{iRe^{i\theta}}-1)d\theta + \int_{-R}^{R} \frac{e^{it}-1}{it}dt$$  ($Re^{[0, \pi]}$ and $t: [-R, R]$ being the respective parametrizations.)  By splitting the left integral in two, I get $\int_{0}^{\pi} e^{iRe^{i\theta}}d\theta \rightarrow 0$ by dominated convergence, and $\int_{0}^{\pi} -1d\theta = -\pi$. By making $R \rightarrow \infty$, I'm left with  $$\pi = \int_{-\infty}^{\infty} \frac{e^{it} - 1}{it}dt = \int_{-\infty}^{\infty}\frac{\sin t}{t}dt + \int_{-\infty}^{\infty}\frac{\cos t -1}{it}dt.$$  How do I get rid of the integral on the right? I welcome all feedback. 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