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NCERT Exemplar · Class 11 Mathematics Complex Numbers and Quadratic Equations

50 questions · 50 still being checked

EXERCISE 5.3 41–50 (part 5 of 5)

  1. Choose the correct answer from the given four options indicated against each of the Exercises from $\displaystyle 35$ to $\displaystyle 50$ (M.C.Q)

    Exercise 41

    Which of the following is correct for any two complex numbers z1\displaystyle z_1 and z2\displaystyle z_2 ?
    (A)
    ∣z1z2∣=∣z1∣∣z2∣\displaystyle \left|z_1 z_2\right|=\left|z_1\right|\left|z_2\right|
    (B)
    arg⁡(z1z2)=arg⁡(z1).arg⁡(z2)\displaystyle \arg \left(z_1 z_2\right)=\arg \left(z_1\right) . \arg \left(z_2\right)
    (C)
    ∣z1+z2∣=∣z1∣+∣z2∣\displaystyle \left|z_1+z_2\right|=\left|z_1\right|+\left|z_2\right|
    (D)
    ∣z1+z2∣≥∣z1∣−∣z2∣\displaystyle \left|z_1+z_2\right| \geq\left|z_1\right|-\left|z_2\right|

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    (A) and (D) both hold for all \(\displaystyle z_1, z_2\); (B) and (C) fail.\[|z_1z_2| = |z_1|\,|z_2| \]\[|z_1| = |(z_1+z_2)+(-z_2)| \le |z_1+z_2| + |z_2| \]\[\Rightarrow\ |z_1+z_2| \ge |z_1|-|z_2| \]\[z_1=1,\ z_2=-1:\ |z_1+z_2| = 0 \ne 2 = |z_1|+|z_2| \quad \text{(C fails)} \]\[z_1=z_2=i:\ \arg(z_1z_2)=\pi \ne \tfrac{\pi^2}{4} = \arg z_1\cdot\arg z_2 \quad \text{(B fails)} \]Answer: (A) and (D) are both true for every pair; (B) and (C) are false.

  2. Exercise 42

    The point represented by the complex number 2−i\displaystyle 2-i is rotated about origin through an angle π2\displaystyle \frac{\pi}{2} in the clockwise direction, the new position of point is:
    (A)
    1+2i\displaystyle 1+2 i
    (B)
    −1−2i\displaystyle -1-2 i
    (C)
    2+i\displaystyle 2+i
    (D)
    −1+2i\displaystyle -1+2 i

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    NCERT’s answer
    B
    (B) \(\displaystyle -1-2i\). Clockwise rotation through \(\displaystyle \frac{\pi}{2}\) multiplies by \(\displaystyle e^{-i\pi/2}=-i\).\[(2-i)(-i) = -2i + i^2 \]\[= -1-2i \]Answer: (B) \(\displaystyle -1-2i\)

  3. Exercise 43

    Let x,y∈R\displaystyle x, y \in \mathbf{R}, then x+iy\displaystyle x+i y is a non real complex number if:
    (A)
    x=0\displaystyle x=0
    (B)
    y=0\displaystyle y=0
    (C)
    x≠0\displaystyle x \neq 0
    (D)
    y≠0\displaystyle y \neq 0

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    NCERT’s answer
    D
    (D) \(\displaystyle y\ne 0\). The number is non-real exactly when its imaginary part is nonzero.\[\operatorname{Im}(x+iy) = y \ne 0 \]Answer: (D) \(\displaystyle y\ne 0\)

  4. Exercise 44

    If a+ib=c+id\displaystyle a+i b=c+i d, then
    (A)
    a2+c2=0\displaystyle a^2+c^2=0
    (B)
    b2+c2=0\displaystyle b^2+c^2=0
    (C)
    b2+d2=0\displaystyle b^2+d^2=0
    (D)
    a2+b2=c2+d2\displaystyle a^2+b^2=c^2+d^2

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    NCERT’s answer
    D
    (D) \(\displaystyle a^2+b^2=c^2+d^2\). Equal complex numbers have equal real and imaginary parts.\[a+ib=c+id \Rightarrow a=c,\ b=d \]\[a^2+b^2 = c^2+d^2 \]Answer: (D) \(\displaystyle a^2+b^2=c^2+d^2\)

  5. Exercise 45

    The complex number z\displaystyle z which satisfies the condition ∣i+zi−z∣=1\displaystyle \left|\frac{i+z}{i-z}\right|=1 lies on
    (A)
    circle x2+y2=1\displaystyle x^2+y^2=1
    (B)
    the x\displaystyle x-axis
    (C)
    the y\displaystyle y-axis
    (D)
    the line x+y=1\displaystyle x+y=1.

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    NCERT’s answer
    B
    (B) the \(\displaystyle x\)-axis. Put \(\displaystyle z=x+iy\); modulus \(\displaystyle 1\) means numerator and denominator have equal modulus.\[|i+z| = |i-z| \]\[x^2+(y+1)^2 = x^2+(1-y)^2 \]\[4y=0 \Rightarrow y=0 \]Answer: (B) the \(\displaystyle x\)-axis

  6. Exercise 46

    If z\displaystyle z is a complex number, then
    (A)
    ∣z2∣>∣z∣2\displaystyle \left|z^2\right|>|z|^2
    (B)
    ∣z2∣=∣z∣2\displaystyle \left|z^2\right|=|z|^2
    (C)
    ∣z2∣<∣z∣2\displaystyle \left|z^2\right|<|z|^2
    (D)
    ∣z2∣≥∣z∣2\displaystyle \left|z^2\right| \geq|z|^2

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    NCERT’s answer
    B
    (B) \(\displaystyle |z^2|=|z|^2\). The modulus of a product is the product of the moduli.\[|z^2| = |z\cdot z| = |z|\,|z| = |z|^2 \]Answer: (B) \(\displaystyle |z^2|=|z|^2\)

  7. Exercise 47

    ∣z1+z2∣=∣z1∣+∣z2∣\displaystyle \left|z_1+z_2\right|=\left|z_1\right|+\left|z_2\right| is possible if
    (A)
    z2=zˉ1\displaystyle z_2=\bar{z}_1
    (B)
    z2=1z1\displaystyle z_2=\frac{1}{z_1}
    (C)
    arg⁡(z1)=arg⁡(z2)\displaystyle \arg \left(z_1\right)=\arg \left(z_2\right)
    (D)
    ∣z1∣=∣z2∣\displaystyle \left|z_1\right|=\left|z_2\right|

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    NCERT’s answer
    C
    (C) \(\displaystyle \arg z_1=\arg z_2\). Equality holds when both numbers point the same way: \(\displaystyle z_1=r_1e^{i\theta}\), \(\displaystyle z_2=r_2e^{i\theta}\).\[|z_1+z_2| = |(r_1+r_2)e^{i\theta}| = r_1+r_2 = |z_1|+|z_2| \]The other conditions do not force it:\[z_1=i,\ z_2=-i:\ |z_1+z_2| = 0 \ne 2 = |z_1|+|z_2| \]This pair satisfies (A), (B) and (D), yet equality fails.Answer: (C) \(\displaystyle \arg z_1=\arg z_2\)

  8. Exercise 48

    The real value of θ\displaystyle \theta for which the expression 1+icos⁡θ1−2icos⁡θ\displaystyle \frac{1+i \cos \theta}{1-2 i \cos \theta} is a real number is:
    (A)
    nπ+π4\displaystyle n \pi+\frac{\pi}{4}
    (B)
    nπ+(−1)nπ4\displaystyle n \pi+(-1)^n \frac{\pi}{4}
    (C)
    2nπ±π2\displaystyle 2 n \pi \pm \frac{\pi}{2}
    (D)
    none of these.

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    NCERT’s answer
    C
    (C) \(\displaystyle 2n\pi\pm\frac{\pi}{2}\). Rationalise and set the imaginary part to zero.\[\frac{1+i\cos\theta}{1-2i\cos\theta}\cdot\frac{1+2i\cos\theta}{1+2i\cos\theta} = \frac{1-2\cos^2\theta+3i\cos\theta}{1+4\cos^2\theta} \]\[\operatorname{Im} = \frac{3\cos\theta}{1+4\cos^2\theta} = 0 \Rightarrow \cos\theta = 0 \]\[\theta = 2n\pi \pm \frac{\pi}{2} \]Answer: (C) \(\displaystyle 2n\pi\pm\frac{\pi}{2}\)

  9. Exercise 49

    The value of arg⁡(x)\displaystyle \arg (x) when x<0\displaystyle x<0 is:
    (A)
    0\displaystyle 0 (B) π2\displaystyle \frac{\pi}{2}
    (C)
    π\displaystyle \pi
    (D)
    none of these

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    NCERT’s answer
    C
    (C) \(\displaystyle \pi\). A negative real number lies on the negative real axis, so its argument is \(\displaystyle \pi\). \[x = -|x| = |x|(\cos\pi + i\sin\pi) \] \[\arg(x) = \pi \]

  10. Exercise 50

    If f(z)=7−z1−z2\displaystyle f(z)=\frac{7-z}{1-z^2}, where z=1+2i\displaystyle z=1+2 i, then ∣f(z)∣\displaystyle |f(z)| is
    (A)
    ∣z∣2\displaystyle \frac{|z|}{2}
    (B)
    ∣z∣\displaystyle |z| (C) 2∣z∣\displaystyle 2|z|
    (D)
    none of these.

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    NCERT’s answer
    A
    (A) \(\displaystyle \dfrac{|z|}{2}\). \[z^2 = (1+2i)^2 = 1 + 4i - 4 = -3+4i \] \[1 - z^2 = 4 - 4i, \qquad 7 - z = 6 - 2i \] \[f(z) = \frac{6-2i}{4-4i} = \frac{3-i}{2-2i}\cdot\frac{2+2i}{2+2i} = \frac{6+6i-2i+2}{8} = 1 + \frac{i}{2} \] \[|f(z)| = \sqrt{1 + \tfrac14} = \frac{\sqrt5}{2} \] \[|z| = \sqrt{1^2+2^2} = \sqrt5 \;\Rightarrow\; |f(z)| = \frac{|z|}{2} \]