The imaginary part of z -1cos- i sin- z -1-1-cos- i sin- is zero- ifA- - z -1-1B- z -1-2 -C- z -1-D- - z -1-2

Let z 1 = r(cosθ + isinθ) = nbsp; re^iθ ∴ Given expression =re^iθ· e^ iα + 1re^iθe^iα = re^i(θ α) + 1re^ i(θ α) Since, imaginary part of given expressio ...

You visited us 9 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Geometrical Representation of Argument and Modulus

The imaginary...

Question

The imaginary part of $(z - 1) (cos α - i sin α) + (z - 1)^{- 1} \times (cos α + i sin α)$ is zero, if

| z - 1 | = 1

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

arg (z - 1) = 2 α

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

arg (z - 1) = α

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

| z - 1 | = 2

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

Open in App

Solution

The correct options are
A $| z - 1 | = 1$
C $arg (z - 1) = α$
Let $z - 1 = r (cos θ + i sin θ) = r e^{i θ}$
$∴$ Given expression
$= r e^{i θ} \cdot e^{- i α} + \frac{1}{r e^{i θ}} e^{i α}$
$= r e^{i (θ - α)} + \frac{1}{r} e^{- i (θ - α)}$
Since, imaginary part of given expression is zero, we have
$r sin (θ - α) - \frac{1}{r} sin (θ - α) = 0$
$\Rightarrow sin (θ - α) (r - \frac{1}{r}) = 0$

$\Rightarrow r^{2} = 1$
$\Rightarrow r = 1$
$∴ | z - 1 | = 1$
or $sin (θ - α) = 0 \Rightarrow θ - α = 0$
$\Rightarrow θ = α$
$∴ arg (z - 1) = α$

Suggest Corrections

Similar questions

Q. If the imaginary part of the complex number

(z - 1) (cos α - i sin α) + (z - 1)^{- 1} (cos α + i sin α)

is zero, then

Q. Let

z_{1}

and

z_{2}

be complex numbers such that

z_{1} \neq z_{2}

and

| z_{1} |

| z_{2} |

. If

z_{1}

has positive real part and

z_{2}

has negative imaginary part, then

\frac{z_{1} + z_{2}}{z_{1} - z_{2}}

may be

Q. Let

z_{1}

and

z_{2}

be complex numbers such that

z_{1} \neq z_{2}

and

| z_{1} | = | z_{2} |

. If

z_{1}

has positive real part and

z_{2}

has negative imaginary part, then

\frac{z_{1} + z_{2}}{z_{1} - z_{2}}

may be

Q. If

z

is a complex number not purely real such that imaginary part of

z - 1 + \frac{1}{z - 1}

is zero, then locus of

z

is-

Q. Let

z_{1}

and

z_{2}

be complex number such that

z_{1} \neq z_{2}

and

| z_{1} | = | z_{2} | .

z_{1}

has positive real part and

z_{2}

has negative imaginary part, then

\frac{z_{1} + z_{2}}{z_{1} - z_{2}}

may be

Join BYJU'S Learning Program

The imaginary part of (z−1)(cosα−isinα)+(z−1)−1×(cosα+isinα) is zero, if

The imaginary part of $(z - 1) (cos α - i sin α) + (z - 1)^{- 1} \times (cos α + i sin α)$ is zero, if