Test Index

Complex Numbers Part 2

Section: Mathematics

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Question : 75 of 106

Marks: +1, -0

Let

a \;\; \neq b

be two-zero real numbers. Then the number of elements in the set

X = \{ z \in C : \operatorname{Re} (a z^2 + bz) = a.

\operatorname{Re} (b z^2 + a z) = b \}

[6-Apr-2023 shift 2]

0
2
1
3

Solution:

(1) Bonus

\because z + \overline{z} = 2 \operatorname{Re}(z) \;\; \text{ If } z = x + iy

\Rightarrow z + \overline{z} = 2 x

z^2 + (\overline{z})^2 = 2 (x^2 - y^2)

(a z^2 + b z) + (a \overline{z}^2 + b \overline{z}) = 2 a

. . . (1)

(b z^2 + a z) + (b \overline{z}^2 + a \overline{z}) = 2 b

. . . (2)

\;\; \text{add (1) and (2)} \;

(a+b) z^2 + (a+b) z + (a+b) \overline{z}^2 + (a+b) \overline{z} = 2(a+b)

(a+b) [z^2 + z + (\overline{z})^2 + \overline{z}] = 2(a+b)

\;\; \text{sub. (1) and (2)} \;

(a - b) [z^2 - z + \overline{z}^2 - \overline{z}] = 2(a - b)

. . . (3)

z^2 + \overline{z}^2 - z - \overline{z} = 2

. . . (4)

\;\; \text{Case I: If } a + b \neq 0

\;\; \text{From (3) \& (4)} \;

2 x + 2 (x^2 - y^2) = 2 \Rightarrow x^2 - y^2 + x = 1

. . . (5)

2 (x^2 - y^2) - 2 x = 2 \Rightarrow x^2 - y^2 - x = 1

. . . (6)

(5) - (6)

2 x = 0 \Rightarrow x = 0

\;\; \text{from (5)} \; y^2 = -1 \;\; \Rightarrow \; \text{not possbible} \;

\therefore \; \text{Ans} \; = 0

Case II: If

a+b=0

then infinite number of solution.

So, the set

X

have infinite number of elements.

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