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The range of real number for which the equation z + |z - 1|

Question

The range of real number $α$ for which the equation $z + α ∣ z - 1∣ + 2 i = 0$ has a solution is

[- \frac{5}{2}, \frac{5}{2}]

[- \frac{3}{2}, \frac{3}{2}]

[0, \frac{5}{2}]

(- \infty, - \frac{5}{2}] \cup [\frac{5}{2}, \infty)

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Eleanor Discussed

The range of real number

α

for which the equation

z + α ∣ z - 1∣ + 2 i = 0

has a solution is

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

We have,

z + α ∣ z - 1∣ + 2 i = 0 \Rightarrow x + i (y + 2) + α (x - 1)^{2} + y^{2} = 0

Equating real and imaginary parts

y + 2 = 0 \Rightarrow y = - 2

and

x + α (x - 1)^{2} + 4 = 0

∴ x^{2} = α^{2} (x^{2} - 2 x + 5)

\Rightarrow (1 - α^{2}) x^{2} + 2 α^{2} x - 5 α^{2} = 0

Since

x

is real,

∴ D = B^{2} - 4 A C \geq 0 \Rightarrow 4 α^{4} + 20 α^{2} (1 - α^{2}) \geq 0

\Rightarrow - 4 α^{4} + 5 α^{2} \geq 0 \Rightarrow 4 α^{2} (α^{2} - \frac{5}{4}) \leq 0

Now

α

being real implies

α^{2}

is +ve and hence we conclude that

α^{2} - \frac{5}{4}

i -ve

\Rightarrow - \frac{5}{2} \leq α \leq \frac{5}{2}

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Question Text	The range of real number $α$ for which the equation $z + α ∣ z - 1∣ + 2 i = 0$ has a solution is
Answer Type	Text solution:1
Upvotes	150