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Complex Numbers and Quadratic Equations — JEE MCQs with answers

NCERT Class 11 · Mathematics · Ch 410 free questions563 in the full bank

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Question 1 · 4.4 The Modulus and the Conjugate of a Complex Number · Easy
For a complex number $z = a + ib$, the relation $z\,\overline{z} = |z|^2$ says that $z\,\overline{z}$ equals
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Answer: B — Option B is correct. Since $|z| = \sqrt{a^2 + b^2}$, squaring gives $|z|^2 = a^2 + b^2$, and the page states $z\,\overline{z} = |z|^2$. The value $a^2 - b^2$ takes $i^2 = +1$ in $(ib)(-ib)$; $a^2 + b^2 + 2abi$ adds the cross terms $-iab$ and $+iab$ instead of letting them cancel; and $a^2 - b^2 + 2abi$ multiplies $z$ by itself instead of by $\overline{z}$.

NCERT Class 11 Mathematics · Ch 4 · §4.4 ("Observe that the multiplicative inverse of the non-zero complex number z is given by")

Question 2 · 4.3 Algebra of Complex Numbers · Medium
Express $(5-3i)^3$ in the form $a+ib$.
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Answer: C — This is Example 3. By $(z_1-z_2)^3 = z_1^3 - 3z_1^2z_2 + 3z_1z_2^2 - z_2^3$ with $z_1=5$, $z_2=3i$: $125-3(25)(3i)+3(5)(3i)^2-(3i)^3 = 125-225i-135+27i = -10-198i$. Option C is correct because it is this value. The value $260 - 252i$ takes $i^2 = +1$ (and so $i^3 = +i$); $-10 + 27i$ leaves out the $-3z_1^2z_2$ term; and $260 - 27i$ takes $i^2 = +1$ (and so $i^3 = +i$), and also leaves out the $-3z_1^2z_2$ term.

NCERT Class 11 Mathematics · Ch 4 · §4.3 ("Example 3 Express (5 - 3i)3 in the form a + ib.")

Question 3 · 4.2 Complex Numbers · Hard
If $3x + i(2x + y) = 9 + i11$, where $x$ and $y$ are real numbers, then $y$ equals
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Answer: B — Option B is correct: equating real parts gives $3x = 9$, so $x = 3$; equating imaginary parts gives $2x + y = 11$, so $6 + y = 11$ and $y = 5$. The value $11$ mistakes the imaginary part of the right-hand side for $y$. The value $17$ comes from $2x + y = 11$ solved as $y = 11 + 6$. The value $9$ is the real part of the right-hand side.

NCERT Class 11 Mathematics · Ch 4 · §4.2 ("Two complex numbers z1 = a + ib and z2 = c + id are equal if a = c and b = d.")

Question 4 · 4.5 Argand Plane and Polar Representation · Hard
Example 8 asks to prove a result for $x + iy = \dfrac{a+ib}{a-ib}$. The value of $x^{2}+y^{2}$ is:
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Answer: A — Option A is correct because $x^{2}+y^{2} = \dfrac{(a^{2}-b^{2})^{2}+4a^{2}b^{2}}{(a^{2}+b^{2})^{2}} = \dfrac{(a^{2}+b^{2})^{2}}{(a^{2}+b^{2})^{2}} = 1$, the very result the example establishes. The value is a pure number, so it cannot depend on $a$ and $b$.

NCERT Class 11 Mathematics · Ch 4 · §4.5 (", prove that x2 + y2 = 1.")

Question 5 · 4.1 Introduction · Hard
For the equation $x^2 + kx + 4 = 0$, which of the following values of $k$ makes $D = b^2 - 4ac < 0$?
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Answer: D — Option D is correct. With $a = 1$, $b = k$, $c = 4$, $D = k^2 - 16$. For $k = 2$, $D = 4 - 16 = -12$, which is negative. For $k = 4$, $D = 16 - 16 = 0$; for $k = 5$, $D = 25 - 16 = 9$; for $k = -6$, $D = 36 - 16 = 20$ — squaring makes the sign of $k$ irrelevant, so a negative $k$ of large size does not help.

NCERT Class 11 Mathematics · Ch 4 · §4.1 ("the main objective is to solve the equation ax2 + bx + c = 0, where D = b2 - 4ac < 0")

Question 6 · 4.2 Complex Numbers · Medium
If $(5 - x) + i(y + 3) = 2 + i(-8)$, where $x$ and $y$ are real numbers, then
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Answer: C — Option C is correct: equating real parts gives $5 - x = 2$, so $x = 3$; equating imaginary parts gives $y + 3 = -8$, so $y = -11$. Option D solves $5 + x = 2$ instead of $5 - x = 2$; then $5 - x = 8$. Option A moves the $3$ with the wrong sign, $y = -8 + 3 = -5$; then $y + 3 = -2$, not $-8$. Option B makes both slips.

NCERT Class 11 Mathematics · Ch 4 · §4.2 ("Two complex numbers z1 = a + ib and z2 = c + id are equal if a = c and b = d.")

Question 7 · 4.4 The Modulus and the Conjugate of a Complex Number · Hard
The multiplicative inverse of $1 - \sqrt{2}\,i$ is
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Answer: A — Option A is correct because $a = 1$, $b = -\sqrt{2}$ give $a^{2} + b^{2} = 1 + 2 = 3$, so $z^{-1} = \dfrac{1 + \sqrt{2}\,i}{3}$. The value $\dfrac{1}{3} - \dfrac{\sqrt{2}}{3}i$ keeps $z$ itself in the numerator instead of $\bar z = 1 + \sqrt{2}\,i$; $1 + \sqrt{2}\,i$ leaves out the division by $|z|^{2} = 3$; and $-1 + \sqrt{2}\,i$ keeps $z$ itself in the numerator instead of $\bar z = 1 + \sqrt{2}\,i$, and also divides by $a^{2} - b^{2} = -1$ instead of $a^{2} + b^{2}$.

NCERT Class 11 Mathematics · Ch 4 · §4.4 ("Observe that the multiplicative inverse of the non-zero complex number z is")

Question 8 · 4.3 Algebra of Complex Numbers · Medium
The value of $(2+i)^2+(2-i)^2$ is
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Answer: A — By the two square identities of §4.3.7, $(2+i)^2 = 4-1+4i = 3+4i$ and $(2-i)^2 = 4-1-4i = 3-4i$. Their sum is $6$, the cross terms cancelling. Option A is correct because it is this value. The value $10$ comes from taking $i^2=+1$ in both squares; $6 + 8i$ adds the cross terms instead of cancelling them; $8$ drops the $z_2^2$ term from both squares.

NCERT Class 11 Mathematics · Ch 4 · §4.3 ("= (z1 + z2) z1 + (z1 + z2) z2")

Question 9 · 4.5 Argand Plane and Polar Representation · Medium
§4.5 states that the complex number $x + iy$ can be represented geometrically as which unique point of the XY-plane?
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Answer: D — Option D is correct because the text says $x + iy$ can be represented geometrically as the unique point $P(x, y)$ in the XY-plane. $P(y, x)$ swaps the coordinates, $P(x, -y)$ reverses the sign of the ordinate, and $P(0, x + y)$ collapses both coordinates into a single sum, so none of them is the point named in the text.

NCERT Class 11 Mathematics · Ch 4 · §4.5 ("geometrically as the unique point P(x, y)")

Question 10 · 4.1 Introduction · Medium
For the equation $x^2 + x + 1 = 0$, the value of $D = b^2 - 4ac$ is
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Answer: A — Option A is correct. Here $a = 1$, $b = 1$, $c = 1$, so $D = 1^2 - 4(1)(1) = 1 - 4 = -3$. None of the other values arises from $b^2 - 4ac$ for these coefficients.

NCERT Class 11 Mathematics · Ch 4 · §4.1 ("the main objective is to solve the equation ax2 + bx + c = 0, where D = b2 - 4ac < 0")

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