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- Thread starter jacks
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- Feb 5, 2012

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Hi jacks,If [tex]z[/tex] and [tex] \omega[/tex] are two complex no. such that [tex]\mid z \mid =\mid \omega \mid = 1[/tex] and [tex]\mid z+i\omega \mid = \mid z-i\omega \mid = 2[/tex].Then find value of [tex]z[/tex]

Take, \(z=x_1+iy_1\mbox{ and }w=x_2+iy_2\)

Since, \(\displaystyle \mid z+i\omega \mid = \mid z-i\omega \mid\),

\[\mid(x_1+iy_1)+i(x_2+iy_2)\mid=\mid(x_1+iy_1)-i(x_2+iy_2)\mid\]

\[\Rightarrow\mid(x_1-y_2)+i(x_2+y_1)\mid=\mid(x_1+y_2)+i(y_1-x_2)\mid\]

\[\Rightarrow (x_1-y_2)^2+(x_2+y_1)^2=(x_1+y_2)^2+(y_1-x_2)^2\]

\[\Rightarrow x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}-2x_{1}y_2+2x_{2}y_1=x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}+2x_{1}y_2-2x_{2}y_1\]

\[\Rightarrow -2x_{1}y_2+2x_{2}y_1=2x_{1}y_2-2x_{2}y_1\]

\[\Rightarrow x_{1}y_2-x_{2}y_1=0\]

Since, \(\mid z-i\omega \mid = 2\),

\begin{equation}x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}+2x_{1}y_2-2x_{2}y_1=4\end{equation}

Also, \(\mid z \mid =\mid \omega \mid = 1\Rightarrow x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}=2\)

\[\therefore x_{1}y_2-x_{2}y_1=1\]

This is a contradiction since we have obtained \(x_{1}y_2-x_{2}y_1=0\). There are no complex numbers \(z\) and \(w\) satisfying the given conditions.

- Jan 30, 2012

- 2,541

Since |z| = 1, the only way for |z + iw| to be 2 is for z to coincide with iw, but then |z - iw| = 0.