Question

If the imaginary part of the expression $\frac{z-1}{e^{i\theta }}+\frac{e^{i\theta }}{z-1}$ be zero, then lucus of z is :

• A. st.line
• B. parabola
• C. unit circle
• D. none

Answer 1

The correct option is C unit circle

Given I.P. of $U+\frac{1}{U}=0whereU=\frac{z-1}{e^{i\theta }}.$
Now ,I.P.of $Z=0\Rightarrow Z-\overline{Z}=2iY=0$
$\therefore (U+\frac{1}{U})-\left(\overline{U+\frac{1}{U}}\right)=0$
or $(U+\frac{1}{U})-(\overline{U}+\frac{1}{\overline{U}})=0$
or $(U-\overline{U})+(\frac{1}{U}-\frac{1}{U})=0$
or $(U-\overline{U})(1-\frac{1}{{\left|U\right|}^2})=0$
or ${\left|U\right|}^2=1asU\ne \overline{U}$
${\left|\frac{z-1}{e^{i\theta }}\right|}^2=\frac{{|z-1|}^2}{1}=1as\left|e^{i\theta }\right|=1$
$\therefore |z-1|=1$
which represents a unit circle.