Gujarat BoardEnglish MediumSTD 11 ScienceMATHSComplex Numbers3 Marks
Question
If $\frac{\text{z}-1}{\text{z}+1}$ is purely imaginary number $(\text{z}\neq-1),$ find the value of $|\text{z}|.$
✓
Answer
Let $\text{z}=\text{x}+\text{iy},$ $\frac{\text{z}-1}{\text{z}+1}$ $=\frac{\text{x+iy}-1}{\text{x+iy}+1}$ $=\frac{\text{x-1+iy}}{\text{x+1+iy}}$ $=\frac{(\text{x-1+iy)}(\text{x+1-iy)}}{(\text{x+1+iy)}(\text{x+1-iy)}}$ [Rationalizing the denominator] $=\frac{(\text{x-1+iy)}(\text{x+1-iy)}}{{(\text{x}+1)}^2-{(\text{iy)}^2}}$ $=\frac{\text{x}^2+\text{x}-\text{ixy}-\text{x}-1+\text{iy}+\text{ixy}+\text{iy}+\text{y}^2}{\text{x}^2+2\text{x}+1+\text{y}^2}$ $=\frac{\text{x}^2-1+2\text{iy}+\text{y}^2}{\text{x}^2+2\text{x}+1\text{y}^2}$ $=\frac{\text{x}^2+\text{y}^2-1}{\text{x}^2+2\text{x}+1+\text{y}^2}+\text{i}\frac{2\text{y}}{\text{x}^2+2\text{x}+1+\text{y}^2}$ $\because$ It is a purely imaginary no. therefore real part = 0 $=\frac{\text{x}^2+\text{y}^2-1}{\text{x}^2+2\text{x}+1+\text{y}^2}=0$ $\Rightarrow\text{x}^2+\text{y}^2-1=0$ $\Rightarrow\text{x}^2+\text{y}^2=1$ $\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=1$ $\Rightarrow|\text{z}|=1$
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