You visited us 1 times! Enjoying our articles? Unlock Full Access!

Rate of Change

If z is a com...

Question

If z is a complex number, then $z^{2} + {¯ ¯ ¯ z}^{2} = 2$ represents

a circle

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a straight line

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a hyperbola

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

an ellipse

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C a hyperbola
Let
$z = x + i y$
$¯ z = (x - i y)$
Hence
$(x + i y)^{2} + (x - i y)^{2} = 2$
$2 (x^{2} - y^{2}) = 2$
$x^{2} - y^{2} = 1$
Hence an equation of hyperbola.

Suggest Corrections

Similar questions

z

z^{2} + z | z | + | z |^{2} = 0,

z

If $| z - z_{1} | + | z + z_{2} |$ = $λ, λ \in R^{+}, z_{1}, z_{2}$ are fixed complex numbers, represents an ellipse, then:

z

z^{2} = (¯ ¯ ¯ z)^{2}

{|z|}^{2} > {|z|}^{2}

{|z|}^{2} = {|z|}^{2}

{|z|}^{2} < {|z|}^{2}

{|z|}^{2} \geq {|z|}^{2}

| z - z_{1} | + | z - z_{2} | = k (> | z_{1} - z_{2} |)

z

z_{1}

z_{2}

Join BYJU'S Learning Program