Question

complex number question

Answer 1

\[\left| z^{2}- 1\right|=|z|^{2}+1\], then z lies on

Answer 2

a) the real axis b) the imaginary axis c) a circle d) an elipse

Answer 3

@Michele_Laino @ganeshie8 @ShadowLegendX

Answer 4

@mathmath333

Answer 5

here we can write, this:
\[z = x + iy\]
so we get:
\[\begin{gathered}
{z^2} - 1 = \left( {{x^2} - {y^2} - 1} \right) + 2ixy \hfill \\
\hfill \\
{\left| z \right|^2} + 1 = {x^2} + {y^2} + 1 \hfill \\
\end{gathered} \]

Answer 6

we replace into the original equation, and we get:
\[\sqrt {{{\left( {{x^2} - {y^2} - 1} \right)}^2} + 4{x^2}{y^2}} = {x^2} + {y^2} + 1\]
please square both sides, and simplify, what condition do you get?

Answer 7

4x^2=0

Answer 8

ok i got it so it lies on imaginary axis because x=0
i could do it till there on my own but knnow i got it lies on imaginary axis

Answer 9

thast's right!
so we can write:
\(x=0\), and the complex numbers which satisfy that equation, are:
\[z = iy,\quad y \in \mathbb{R}\]

Answer 10

that's*...

Answer 11

*