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For every real number c> 0, find all complex numbers z, satis

Question

For every real number $c > 0,$ find all complex numbers z, satisfying the equation $z ∣ z ∣ + cz + i = 0.$

(x, y) = (0, c \pm \frac{c ^{2} - 4}{2})

(x, y) = (0, - c \pm \frac{c ^{2} - 4}{2})

(x, y) = (0, c \pm \frac{c ^{2} - 3}{2})

(x, y) = (0, - c \pm \frac{c ^{2} - 3}{2})

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Liam Discussed

For every real number

c > 0,

find all complex numbers z, satisfying the equation

z ∣ z ∣ + cz + i = 0.

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

Let

z = x + i y

Hence ,

(x + i y) (x^{2} + y^{2}) + c x + i cy + i = 0

Hence the real part is

\Rightarrow x x^{2} + y^{2} + c x = 0

Hence ,

x = 0

since

∣ z ∣ \neq = 0

And substituting

x = 0

, we get the imaginary part as

\Rightarrow y^{2} + cy + 1 = 0

y = \frac{- c \pm c ^{2} - 4}{2}

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Question Text	For every real number $c > 0,$ find all complex numbers z, satisfying the equation $z ∣ z ∣ + cz + i = 0.$
Answer Type	Text solution:1
Upvotes	150