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Complex Numbers Part 1

Section: Mathematics

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Question : 92 of 100

Marks: +1, -0

If

\left|z^2-1\right| = |z|^2+1

z

an ellipse
the imaginary axis
a circle
the real axis

Solution:

Given

\left|z^2-1\right| = |z|^2+1

,

By squaring both sides we get,

\left|z^2-1\right|^2 = (|z|^2+1)^2

\Rightarrow (z^2-1) \overline{(z^2-1)} = (|z|^2+1)^2 [.

as

|z|^2 = z \overline{z} ]

\Rightarrow (z^2-1) \left((\overline{z})^2-1\right) = (|z|^2+1)^2

\Rightarrow (z \overline{z})^2 - z^2 - (\overline{z})^2 + 1 = |z|^4 + 2|z|^2 + 1

\Rightarrow |z|^4 - z^2 - (\overline{z})^2 + 1 = |z|^4 + 2|z|^2 + 1

\Rightarrow z^2 + (\overline{z})^2 + 2 z \overline{z} = 0

\Rightarrow (z + \overline{z})^2 = 0

\Rightarrow z + \overline{z} = 0

\Rightarrow z = -\overline{z}

If

z = x + i y

then

\overline{z} = x

-iy

\; \therefore x + i y = -(x - i y)

\; \Rightarrow x + i y = -x + i y

\; \Rightarrow x = 0

\therefore z

is purely imaginary.

So, it is lie on the imaginary axis.

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