Let - be the imaginary part of z - 1 e- i - - z - 1-1 ei - where z is complex and - is real- then - - 0 implies that z lies on a circle of centre - and radius -

The correct option is C centre (1, 0); radius 1 γ=0 implies nbsp; (z 1)^2=e^2iα Or nbsp; |z 1|^2=1 or nbsp; |z 1|=1 Hence if z=x+iy ...

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Standard XII

Mathematics

Domain

Let γ be th...

Question

Let $γ$ be the imaginary part of $(z - 1) e^{- i α} + (z - 1)^{- 1} e^{i α}$ where z is complex and $α$ is real, then $γ = 0$ implies that z lies on a circle of centre .............. and radius ...............

centre (1, 0); radius 2

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centre (-1, 0); radius 2

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centre (1, 0); radius 1

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centre (-1, 0); radius 1

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Solution

The correct option is C centre (1, 0); radius 1
$γ = 0$ implies
$(z - 1)^{2} = e^{2 i α}$
Or
$| z - 1 |^{2} = 1$
or
$| z - 1 | = 1$
Hence if $z = x + i y$
Then $| z - 1 | = 1$ implies
$(x - 1)^{2} + y^{2} = 1$ .
Hence an circle of unit radius with center at $(1, 0)$ .

Suggest Corrections

Similar questions

Q. Let

a, b \in R

and

a^{2} + b^{2} \neq 0

. Suppose

S = {z \in C : z = \frac{1}{a + i b t}, t \in R, t \neq 0},

where

i = \sqrt{- 1}

. If

z = x + i y

and

z \in S,

then

(x, y)

lies on

Q. Let

α, β \in R

and

(1 - cos α)^{2} + {sin}^{2} β \neq 0

. Suppose

S = {z \in C : z = \frac{1}{(1 - cos α) + i k sin β}, k \in R - {0}},

where

i = \sqrt{- 1} .

z = x + i y, z \in S

and

(x, y)

lies on a circle, then

Q. Let

a, b ϵ R

and

a^{2} + b^{2} \neq 0

. Suppose

S = {z ϵ C : z = \frac{1}{a + i b t}, t ϵ R, t \neq 0}

, where

i = \sqrt{- 1}

. If

z = x + i y

and

z ϵ S

, then

(x, y)

lies on

Q. The locus of

z

in the argand diagram is the circle |z|

=

2,

then show that the locus of

| z + 1 |

is circle with centre

(1, 0)

and radius

2.

State True and False

| z - i | < 1

represents all points z inside the circle with centre i.e.(0, 1) and radius 1.

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Let γ be the imaginary part of (z−1)e−iα+(z−1)−1eiα where z is complex and α is real, then γ=0 implies that z lies on a circle of centre .............. and radius ...............

The correct option is C centre (1, 0); radius 1γ=0 implies (z−1)2=e2iαOr |z−1|2=1or |z−1|=1Hence if z=x+iyThen |z−1|=1 implies (x−1)2+y2=1.Hence an circle of unit radius with center at (1,0).

Let $γ$ be the imaginary part of $(z - 1) e^{- i α} + (z - 1)^{- 1} e^{i α}$ where z is complex and $α$ is real, then $γ = 0$ implies that z lies on a circle of centre .............. and radius ...............

The correct option is C centre (1, 0); radius 1
$γ = 0$ implies
$(z - 1)^{2} = e^{2 i α}$
Or
$| z - 1 |^{2} = 1$
or
$| z - 1 | = 1$
Hence if $z = x + i y$
Then $| z - 1 | = 1$ implies
$(x - 1)^{2} + y^{2} = 1$ .
Hence an circle of unit radius with center at $(1, 0)$ .