If I have given that the absolute value of Z is 1. how do I prove the z+1z is re and z−1z is Im?

Step 1 There are two things to note, if |z|=1, then 1z=z′(conjugate of z) and z+z′=2re(z) and z−z′=2im(z) 1z=z′z′z( multiply top and bottom by z′) =z′1=z′ Therefore (z+frac{1}{z}=z+z

Step 1 Make a picture of the situation in the complex plane. You probably know that the absolute value of 1z is also 1, and you should know that the angle that 1z makes with the x axis is the same size but of opposite rotational direction as the angle made by z. It’ll come to you.

Step 1 There are two things to note, if |z|=1, then 1z=z′(conjugate of z) and z+z′=2re(z) and z−z′=2im(z) 1z=z′z′z( multiply top and bottom by z′) =z′1=z′ Therefore (z+frac{1}{z}=z+z

Step 1 Make a picture of the situation in the complex plane. You probably know that the absolute value of 1z is also 1, and you should know that the angle that 1z makes with the x axis is the same size but of opposite rotational direction as the angle made by z. It’ll come to you.

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High School
Algebra
Algebra II
Complex numbers

Answered question

2022-01-17

If I have given that the absolute value of Z is 1. how do I prove the

z + \frac{1}{z}

is re and

z - \frac{1}{z}

is Im?

Answer & Explanation

nick1337

Expert2022-01-17Added 777 answers

Step 1
There are two things to note, if
$| z | = 1$ ,
then
$\frac{1}{z} = z^{'}$ (conjugate of z) and
$z + z^{'} = 2 r e (z)$ and
$z - z^{'} = 2 i m (z)$
$\frac{1}{z} = \frac{z^{'}}{z^{'} z}$ ( multiply top and bottom by $z^{'}$ )
$= \frac{z^{'}}{1} = z^{'}$
Therefore
\(z+\frac{1}{z}=z+z

star233

Skilled2022-01-17Added 403 answers

Step 1
Make a picture of the situation in the complex plane. You probably know that the absolute value of
$\frac{1}{z}$ is also 1,
and you should know that the angle that $\frac{1}{z}$ makes with the x axis is the same size but of opposite rotational direction as the angle made by z.
It’ll come to you.

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