For every real number c≥0, find all the complex numbers z which satisfy the equation 2∣z∣−4cz+1+ic=0.

Let z=x+iy⇒2x2+y2−4c(x+iy)+1+ic=0Considering the imaginary part, we get ⇒−4cy+c=0∴y=41Substituting in the real part, we get ⇒16x2+1−8cx+2=0⇒16x2+1=64c2x2−32cx+4⇒x2(64c2−16)−32cx+3=0⇒x=2(64c2−16)32c±1024c2−768c2+192⇒2(16c2−4)8c±16c2+12⇒4(4c2−1)4c±4c2+3

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For every real number c 0, find all the complex numbers z whi

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For every real number $c \geq 0,$ find all the complex numbers z which satisfy the equation $2 ∣ z ∣ - 4 cz + 1 + i c = 0.$

(x, y) = \frac{4 c \pm 4 c ^{2} + 3}{4 ( 4 c ^{2} - 1 )}, 1/4

(x, y) = \frac{4 c \pm 4 c ^{2} + 3}{4 ( 4 c ^{2} + 1 )}, - 1/4

(x, y) = \frac{4 c \pm 4 c ^{2} + 3}{4 ( 4 c ^{2} - 1 )}, - 1/4

(x, y) = \frac{4 c \pm 4 c ^{2} + 3}{4 ( 4 c ^{2} + 1 )}, 1/4

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For every real number

c \geq 0,

find all the complex numbers z which satisfy the equation

2 ∣ z ∣ - 4 cz + 1 + i c = 0.

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Text solutionVerified

Let

z = x + i y

\Rightarrow 2 x^{2} + y^{2} - 4 c (x + i y) + 1 + i c = 0

Considering the imaginary part, we get

\Rightarrow - 4 cy + c = 0

∴ y = \frac{1}{4}

Substituting in the real part, we get

\Rightarrow 16 x^{2} + 1 - 8 c x + 2 = 0

\Rightarrow 16 x^{2} + 1 = 64 c^{2} x^{2} - 32 c x + 4

\Rightarrow x^{2} (64 c^{2} - 16) - 32 c x + 3 = 0

\Rightarrow x = \frac{32 c \pm 1024 c ^{2} - 768 c ^{2} + 192}{2 ( 64 c ^{2} - 16 )}

\Rightarrow \frac{8 c \pm 16 c ^{2} + 12}{2 ( 16 c ^{2} - 4 )}

\Rightarrow \frac{4 c \pm 4 c ^{2} + 3}{4 ( 4 c ^{2} - 1 )}

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Question Text	For every real number $c \geq 0,$ find all the complex numbers z which satisfy the equation $2 ∣ z ∣ - 4 cz + 1 + i c = 0.$
Answer Type	Text solution:1
Upvotes	150

For every real number c≥0, find all the complex numbers z which satisfy the equation 2∣z∣−4cz+1+ic=0.

Text solutionVerified

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For every real number $c \geq 0,$ find all the complex numbers z which satisfy the equation $2 ∣ z ∣ - 4 cz + 1 + i c = 0.$