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Let {z}={x}+ iy be a complex number where {x} and {y} are inte

Question

Let $z = x + i y$ be a complex number where $x$ and $y$ are integers. Then the area of the rectangle whose vertices are the roots of the equation $\overline{z} z^{3} + z \overline{z}^{3} = 350$ is

48

32

40

80

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Samuel Discussed

Let

z = x + i y

be a complex number where

x

and

y

are integers. Then the area of the rectangle whose vertices are the roots of the equation

\overline{z} z^{3} + z \overline{z}^{3} = 350

11 mins ago

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11 mins ago

Text solutionVerified

\overline{z} z^{3} + z \overline{z}^{3} = 350

z \overline{z} (\overline{z}^{2} + z^{2}) = 350

Put

z = x + i y

(x^{2} + y^{2}) (x^{2} - y^{2}) = 175

Since x and y are integer so only possibility is

x^{2} + y^{2} = 25

and

x^{2} - y^{2} = 7

so solving these equation we get

x = \pm 4, y = \pm 3

Thus required area

= 8 \times 6 = 48 s q . u ni t .

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Question Text	Let $z = x + i y$ be a complex number where $x$ and $y$ are integers. Then the area of the rectangle whose vertices are the roots of the equation $\overline{z} z^{3} + z \overline{z}^{3} = 350$ is
Answer Type	Text solution:1
Upvotes	150