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Question Number 177702 by lapache last updated on 08/Oct/22

Solve in C   { ((x+y=z)),(((1/x)+(1/y)=(1/z))) :}

$${Solve}\:{in}\:\mathbb{C} \\ $$$$\begin{cases}{{x}+{y}={z}}\\{\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{1}}{{z}}}\end{cases} \\ $$

Answered by Frix last updated on 08/Oct/22

(1) x=z−y  (2) x=((yz)/(y−z))  ⇔  z−y=((yz)/(y−z))  y^2 +yz−z^2 =0 ⇒ y=z×((1/2)±((√3)/2)i)  ⇒  x=z×((1/2)∓((√3)/2)i)  z∈C ⇒ z=p+qi   ((x),(y),(z) ) = ((((p+qi)((1/2)±((√3)/2)i))),(((p+qi)((1/2)∓((√3)/2)i))),((p+qi)) )

$$\left(\mathrm{1}\right)\:{x}={z}−{y} \\ $$$$\left(\mathrm{2}\right)\:{x}=\frac{{yz}}{{y}−{z}} \\ $$$$\Leftrightarrow \\ $$$${z}−{y}=\frac{{yz}}{{y}−{z}} \\ $$$${y}^{\mathrm{2}} +{yz}−{z}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:{y}={z}×\left(\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right) \\ $$$$\Rightarrow \\ $$$${x}={z}×\left(\frac{\mathrm{1}}{\mathrm{2}}\mp\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right) \\ $$$${z}\in\mathbb{C}\:\Rightarrow\:{z}={p}+{q}\mathrm{i} \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\\{{z}}\end{pmatrix}\:=\begin{pmatrix}{\left({p}+{q}\mathrm{i}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right)}\\{\left({p}+{q}\mathrm{i}\right)\left(\frac{\mathrm{1}}{\mathrm{2}}\mp\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right)}\\{{p}+{q}\mathrm{i}}\end{pmatrix} \\ $$

Commented by Tawa11 last updated on 08/Oct/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 08/Oct/22

((x+y)/(xy))=(1/z)  ⇒xy=(x+y)z=z^2   x+y=z  ⇒x,y are roots of t^2 −zt+z^2 =0  x, y=((z±(√(z^2 −4z^2 )))/2)=(((1±3i)z)/2)

$$\frac{{x}+{y}}{{xy}}=\frac{\mathrm{1}}{{z}} \\ $$$$\Rightarrow{xy}=\left({x}+{y}\right){z}={z}^{\mathrm{2}} \\ $$$${x}+{y}={z} \\ $$$$\Rightarrow{x},{y}\:{are}\:{roots}\:{of}\:{t}^{\mathrm{2}} −{zt}+{z}^{\mathrm{2}} =\mathrm{0} \\ $$$${x},\:{y}=\frac{{z}\pm\sqrt{{z}^{\mathrm{2}} −\mathrm{4}{z}^{\mathrm{2}} }}{\mathrm{2}}=\frac{\left(\mathrm{1}\pm\mathrm{3}{i}\right){z}}{\mathrm{2}} \\ $$

Commented by Tawa11 last updated on 08/Oct/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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