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NCERT Exemplar · Class 11 Mathematics Complex Numbers and Quadratic Equations

50 questions · 50 still being checked

EXERCISE 5.3 11–20 (part 2 of 5)

  1. Exercise 11

    Solve the equation ∣z∣=z+1+2i\displaystyle |z|=z+1+2 i.

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    NCERT’s answer
    \(\displaystyle \frac{3}{2}-2 i\)
    With \(\displaystyle z=x+iy\): \[\sqrt{x^2+y^2}=(x+1)+i(y+2) \] Imaginary parts: \[y+2=0 \Rightarrow y=-2 \] Real parts: \[\sqrt{x^2+4}=x+1 \] \[x^2+4=x^2+2x+1 \Rightarrow x=\tfrac32 \] Check: \(\displaystyle x+1=\tfrac52>0\) and \(\displaystyle \sqrt{\tfrac94+4}=\tfrac52\).Answer: \(\displaystyle z=\tfrac32-2i\)

  2. Exercise 12

    If ∣z+1∣=z+2(1+i)\displaystyle |z+1|=z+2(1+i), then find z\displaystyle z.

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    NCERT’s answer
    \(\displaystyle \frac{1}{2}-2 i\)
    With \(\displaystyle z=x+iy\): \[\sqrt{(x+1)^2+y^2}=(x+2)+i(y+2) \] Imaginary parts: \[y+2=0 \Rightarrow y=-2 \] Real parts: \[\sqrt{(x+1)^2+4}=x+2 \] \[x^2+2x+5=x^2+4x+4 \Rightarrow x=\tfrac12 \] Check: \(\displaystyle x+2=\tfrac52>0\) and \(\displaystyle \sqrt{\tfrac94+4}=\tfrac52\).Answer: \(\displaystyle z=\tfrac12-2i\)

  3. Exercise 13

    If arg⁡(z−1)=arg⁡(z+3i)\displaystyle \arg (z-1)=\arg (z+3 i), then find x−1:y\displaystyle x-1: y. where z=x+iy\displaystyle z=x+i y

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    NCERT’s answer
    $\displaystyle 1$:$\displaystyle 3$
    Equal arguments give equal tangents. For \(\displaystyle z=x+iy\), \(\displaystyle z-1=(x-1)+iy\) and \(\displaystyle z+3i=x+i(y+3)\): \[\frac{y}{x-1}=\frac{y+3}{x} \] \[xy=(x-1)(y+3)=xy+3x-y-3 \] \[y=3(x-1) \] \[\frac{x-1}{y}=\frac13 \]Answer: \(\displaystyle (x-1):y=1:3\)

  4. Exercise 14

    Show that ∣z−2z−3∣=2\displaystyle \left|\frac{z-2}{z-3}\right|=2 represents a circle. Find its centre and radius.

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    NCERT’s answer
    \(\displaystyle \left(\frac{10}{3}, 0\right), \frac{2}{3}\)
    With \(\displaystyle z=x+iy\): \[|z-2|=2|z-3| \] \[(x-2)^2+y^2=4\left[(x-3)^2+y^2\right] \] \[3x^2+3y^2-20x+32=0 \] \[\left(x-\tfrac{10}{3}\right)^2+y^2=\tfrac{100}{9}-\tfrac{32}{3}=\tfrac49 \] NCERT_Solution_Class11_Maths_Exemplar_Ch5_Ex5-3_Q14 Answer: a circle with centre \(\displaystyle \left(\tfrac{10}{3},0\right)\) and radius \(\displaystyle \tfrac23\).

  5. Exercise 15

    If z−1z+1\displaystyle \frac{z-1}{z+1} is a purely imaginary number (z≠−1)\displaystyle (z \neq-1), then find the value of ∣z∣\displaystyle |z|.

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    NCERT’s answer
    $\displaystyle 1$
    A purely imaginary \(\displaystyle w\) satisfies \(\displaystyle w+\bar w=0\): \[\frac{z-1}{z+1}+\frac{\bar z-1}{\bar z+1}=0 \] \[(z-1)(\bar z+1)+(\bar z-1)(z+1)=0 \] \[2z\bar z-2=0 \] \[|z|^2=1 \]Answer: \(\displaystyle |z|=1\)

  6. Exercise 16

    z1\displaystyle z_1 and z2\displaystyle z_2 are two complex numbers such that ∣z1∣=∣z2∣\displaystyle \left|z_1\right|=\left|z_2\right| and arg⁡(z1)+arg⁡(z2)=\displaystyle \arg \left(z_1\right)+\arg \left(z_2\right)= π\displaystyle \pi, then show that z1=−zˉ2\displaystyle z_1=-\bar{z}_2.

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    Let \(\displaystyle |z_1|=|z_2|=r\) and \(\displaystyle \arg z_1=\theta\). Then \(\displaystyle \arg z_2=\pi-\theta\): \[z_1=r(\cos\theta+i\sin\theta) \] \[z_2=r\left[\cos(\pi-\theta)+i\sin(\pi-\theta)\right]=r(-\cos\theta+i\sin\theta) \] \[\bar z_2=r(-\cos\theta-i\sin\theta)=-r(\cos\theta+i\sin\theta)=-z_1 \]Answer: \(\displaystyle z_1=-\bar z_2\)

  7. Exercise 17

    If ∣z1∣=1(z1≠−1)\displaystyle \left|z_1\right|=1\left(z_1 \neq-1\right) and z2=z1−1z1+1\displaystyle z_2=\frac{z_1-1}{z_1+1}, then show that the real part of z2\displaystyle z_2 is zero.

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    Since \(\displaystyle |z_1|=1\), \(\displaystyle z_1\bar z_1=1\); \(\displaystyle z_1\ne-1\) keeps the denominator nonzero. Multiply by the conjugate of the denominator. \[z_2=\frac{z_1-1}{z_1+1}\cdot\frac{\bar z_1+1}{\bar z_1+1}=\frac{z_1\bar z_1+z_1-\bar z_1-1}{|z_1+1|^2} \] \[z_2=\frac{z_1-\bar z_1}{|z_1+1|^2}\quad (z_1\bar z_1=1) \] \[z_1-\bar z_1=2i\,\operatorname{Im}(z_1) \] \[z_2=\frac{2\operatorname{Im}(z_1)}{|z_1+1|^2}\,i \] Answer: \(\displaystyle \operatorname{Re}(z_2)=0\)

  8. Exercise 18

    If z1,z2\displaystyle z_1, z_2 and z3,z4\displaystyle z_3, z_4 are two pairs of conjugate complex numbers, then find arg⁡(z1z4)+arg⁡(z2z3)\displaystyle \arg \left(\frac{z_1}{z_4}\right)+\arg \left(\frac{z_2}{z_3}\right).

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    NCERT’s answer
    $\displaystyle 0$
    Conjugate pairs: \(\displaystyle z_2=\bar z_1\), \(\displaystyle z_4=\bar z_3\). Let \(\displaystyle w=\dfrac{z_1}{z_4}\). \[\frac{z_2}{z_3}=\frac{\bar z_1}{z_3}=\overline{\left(\frac{z_1}{\bar z_3}\right)}=\overline{\left(\frac{z_1}{z_4}\right)}=\bar w \] \[\arg(\bar w)=-\arg(w) \] \[\arg\!\left(\frac{z_1}{z_4}\right)+\arg\!\left(\frac{z_2}{z_3}\right)=\arg w-\arg w=0 \] Answer: \(\displaystyle 0\) (taking the arguments as \(\displaystyle \theta\) and \(\displaystyle -\theta\), i.e. modulo \(\displaystyle 2\pi\))

  9. Exercise 19

    If ∣z1∣=∣z2∣=…=∣zn∣=1\displaystyle \left|z_1\right|=\left|z_2\right|=\ldots=\left|z_n\right|=1, then show that ∣z1+z2+z3+…+zn∣=∣1z1+1z2+1z3+…+1zn∣\displaystyle \left|z_1+z_2+z_3+\ldots+z_n\right|=\left|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}+\ldots+\frac{1}{z_n}\right|.

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Each \(\displaystyle z_k\) has modulus 1. \[z_k\bar z_k=|z_k|^2=1\;\Rightarrow\;\frac1{z_k}=\bar z_k \] \[\frac1{z_1}+\frac1{z_2}+\dots+\frac1{z_n}=\bar z_1+\bar z_2+\dots+\bar z_n \] \[\bar z_1+\dots+\bar z_n=\overline{z_1+z_2+\dots+z_n} \] \[|\bar w|=|w| \] \[\left|\frac1{z_1}+\dots+\frac1{z_n}\right|=\left|\overline{z_1+\dots+z_n}\right|=|z_1+z_2+\dots+z_n| \] Answer: \(\displaystyle |z_1+\dots+z_n|=\left|\dfrac1{z_1}+\dots+\dfrac1{z_n}\right|\)

  10. Exercise 20

    If for complex numbers z1\displaystyle z_1 and z2\displaystyle z_2, arg (z1)−arg⁡(z2)=0\displaystyle \left(z_1\right)-\arg \left(z_2\right)=0, then show that ∣z1−z2∣=∣z1∣−∣z2∣\displaystyle \left|z_1-z_2\right|=\left|z_1\right|-\left|z_2\right|

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Let \(\displaystyle \arg z_1=\arg z_2=\theta\), \(\displaystyle r_1=|z_1|\), \(\displaystyle r_2=|z_2|\). \[z_1=r_1(\cos\theta+i\sin\theta),\qquad z_2=r_2(\cos\theta+i\sin\theta) \] \[z_1-z_2=(r_1-r_2)(\cos\theta+i\sin\theta) \] \[|z_1-z_2|=|r_1-r_2|\,\sqrt{\cos^2\theta+\sin^2\theta}=|r_1-r_2| \] \[|z_1-z_2|=r_1-r_2=|z_1|-|z_2|\quad (r_1\ge r_2) \] The statement takes \(\displaystyle |z_1|\ge|z_2|\); if \(\displaystyle |z_1|<|z_2|\), swap the roles and the result is \(\displaystyle |z_2|-|z_1|\). Answer: \(\displaystyle |z_1-z_2|=|z_1|-|z_2|\)