Search any question & find its solution
Question:
Answered & Verified by Expert
If $z=x+i y$ is a complex number and $|1+i z|=|1-i z|$, then
Options:
Solution:
1989 Upvotes
Verified Answer
The correct answer is:
$z=\bar{Z}$
We have,
$$
\begin{aligned}
|1+i z| & =|1-i z| \\
\Rightarrow \quad|1+i(x+i y)| & =|1-i(x+i y)| \\
\Rightarrow \quad|(1-y)+i x| & =|(1+y)-i x| \\
\Rightarrow \quad \sqrt{(1-y)^2+x^2} & =\sqrt{(1+y)^2+x^2} \\
\Rightarrow \quad(1-y)^2+x^2 & =(1+y)^2+x^2
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad 1+y^2-2 y=1+y^2+2 y \\
& \Rightarrow \quad 4 y=0 \Rightarrow y=0 \\
& \therefore \quad z=x+i y=x \Rightarrow \bar{z}=x \\
& \therefore \quad z=\bar{z} \\
&
\end{aligned}
$$
$$
\begin{aligned}
|1+i z| & =|1-i z| \\
\Rightarrow \quad|1+i(x+i y)| & =|1-i(x+i y)| \\
\Rightarrow \quad|(1-y)+i x| & =|(1+y)-i x| \\
\Rightarrow \quad \sqrt{(1-y)^2+x^2} & =\sqrt{(1+y)^2+x^2} \\
\Rightarrow \quad(1-y)^2+x^2 & =(1+y)^2+x^2
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad 1+y^2-2 y=1+y^2+2 y \\
& \Rightarrow \quad 4 y=0 \Rightarrow y=0 \\
& \therefore \quad z=x+i y=x \Rightarrow \bar{z}=x \\
& \therefore \quad z=\bar{z} \\
&
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.