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NCERT Exemplar · Class 11 Mathematics Complex Numbers and Quadratic Equations

50 questions · 50 still being checked

EXERCISE 5.3 1–10 (part 1 of 5)

  1. Exercise 1

    For a positive integer n\displaystyle n, find the value of (1−i)n(1−1i)n\displaystyle (1-i)^n\left(1-\frac{1}{i}\right)^n

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    NCERT’s answer
    \(\displaystyle 2^{n}\)
    \[\frac{1}{i} = \frac{i}{i^2} = -i \quad\Rightarrow\quad 1-\frac{1}{i} = 1+i \] \[(1-i)^n(1+i)^n = \big[(1-i)(1+i)\big]^n \] \[(1-i)(1+i) = 1-i^2 = 2 \] \[(1-i)^n\left(1-\frac1i\right)^n = 2^n \] Answer: \(\displaystyle 2^n\)

  2. Exercise 2

    Evaluate ∑n=113(in+in+1)\displaystyle \sum_{n=1}^{13}\left(i^n+i^{n+1}\right), where n∈N\displaystyle n \in \mathbf{N}.

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    NCERT’s answer
    \(\displaystyle -1+i\)
    \[\sum_{n=1}^{13}\left(i^n+i^{n+1}\right) = (1+i)\sum_{n=1}^{13} i^n \] \[i+i^2+i^3+i^4 = i-1-i+1 = 0 \] Twelve terms make three complete cycles: \[\sum_{n=1}^{12} i^n = 0 \quad\Rightarrow\quad \sum_{n=1}^{13} i^n = i^{13} = i \] \[(1+i)\,i = i+i^2 = -1+i \] Answer: \(\displaystyle -1+i\)

  3. Exercise 3

    If (1+i1−i)3−(1−i1+i)3=x+iy\displaystyle \left(\frac{1+i}{1-i}\right)^3-\left(\frac{1-i}{1+i}\right)^3=x+i y, then find (x,y)\displaystyle (x, y).

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    NCERT’s answer
    \(\displaystyle (0, -2)\)
    \[\frac{1+i}{1-i} = \frac{(1+i)^2}{(1-i)(1+i)} = \frac{2i}{2} = i \] \[\frac{1-i}{1+i} = \frac{1}{i} = -i \] \[i^3 = -i, \qquad (-i)^3 = -i^3 = i \] \[i^3-(-i)^3 = -i-i = -2i \] \[x+iy = 0-2i \] Answer: \(\displaystyle (x,y)=(0,\,-2)\)

  4. Exercise 4

    If (1+i)22−i=x+iy\displaystyle \frac{(1+i)^2}{2-i}=x+i y, then find the value of x+y\displaystyle x+y.

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    NCERT’s answer
    \(\displaystyle \frac{2}{5}\)
    \[(1+i)^2 = 1+2i+i^2 = 2i \] \[\frac{2i}{2-i} = \frac{2i(2+i)}{(2-i)(2+i)} = \frac{4i+2i^2}{4-i^2} = \frac{-2+4i}{5} \] \[x=-\frac25, \qquad y=\frac45 \] \[x+y = -\frac25+\frac45 = \frac25 \] Answer: \(\displaystyle \dfrac{2}{5}\)

  5. Exercise 5

    If (1−i1+i)100=a+ib\displaystyle \left(\frac{1-i}{1+i}\right)^{100}=a+i b, then find (a,b)\displaystyle (a, b).

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    NCERT’s answer
    \(\displaystyle (1, 0)\)
    \[\frac{1-i}{1+i} = \frac{(1-i)^2}{(1+i)(1-i)} = \frac{1-2i+i^2}{2} = \frac{-2i}{2} = -i \] The power $\displaystyle 100$ is even, so \[(-i)^{100} = i^{100} = (i^4)^{25} = 1 \] \[a+ib = 1+0\,i \] Answer: \(\displaystyle (a,b)=(1,\,0)\)

  6. Exercise 6

    If a=cos⁡θ+isin⁡θ\displaystyle a=\cos \theta+i \sin \theta, find the value of 1+a1−a\displaystyle \frac{1+a}{1-a}.

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    NCERT’s answer
    \(\displaystyle i \cot \frac{\theta}{2}\)
    Since \(\displaystyle |a|=1\), \(\displaystyle a\bar a=1\) and \(\displaystyle a+\bar a=2\cos\theta\). \[\frac{1+a}{1-a} = \frac{(1+a)(1-\bar a)}{(1-a)(1-\bar a)} \] \[(1+a)(1-\bar a) = 1-\bar a+a-a\bar a = a-\bar a = 2i\sin\theta \] \[(1-a)(1-\bar a) = 1-(a+\bar a)+a\bar a = 2-2\cos\theta \] \[\frac{1+a}{1-a} = \frac{2i\sin\theta}{2(1-\cos\theta)} = \frac{4i\sin\frac\theta2\cos\frac\theta2}{4\sin^2\frac\theta2} = i\cot\frac\theta2 \] Answer: \(\displaystyle i\cot\dfrac{\theta}{2}\)

  7. Exercise 7

    If (1+i)z=(1−i)zˉ\displaystyle (1+i) z=(1-i) \bar{z}, then show that z=−izˉ\displaystyle z=-i \bar{z}.

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    \[(1+i)z = (1-i)\bar z \quad \text{(given)} \] \[z = \frac{1-i}{1+i}\,\bar z \quad \text{(} 1+i\neq 0 \text{)} \] \[\frac{1-i}{1+i} = \frac{(1-i)^2}{(1+i)(1-i)} = \frac{-2i}{2} = -i \] \[z = -i\,\bar z \] Answer: \(\displaystyle z=-i\bar z\)

  8. Exercise 8

    If z=x+iy\displaystyle z=x+i y, then show that zzˉ+2(z+zˉ)+b=0\displaystyle z \bar{z}+2(z+\bar{z})+b=0, where b∈R\displaystyle b \in \mathbf{R}, represents a circle.

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    \[z\bar z=(x+iy)(x-iy)=x^2+y^2, \qquad z+\bar z=2x \] \[x^2+y^2+2(2x)+b=0 \] \[(x^2+4x+4)+y^2=4-b \] \[(x+2)^2+y^2=4-b \] For \(\displaystyle b<4\) this is a circle with centre \(\displaystyle (-2,0)\) and radius \(\displaystyle \sqrt{4-b}\) (\(\displaystyle b=4\) gives only the point \(\displaystyle (-2,0)\); \(\displaystyle b>4\), no point). NCERT_Solution_Class11_Maths_Exemplar_Ch5_Ex5-3_Q8 Answer: a circle with centre \(\displaystyle (-2,0)\) and radius \(\displaystyle \sqrt{4-b}\), for \(\displaystyle b<4\)

  9. Exercise 9

    If the real part of zˉ+2zˉ−1\displaystyle \frac{\bar{z}+2}{\bar{z}-1} is 4\displaystyle 4, then show that the locus of the point representing z\displaystyle z in the complex plane is a circle.

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

    With \(\displaystyle z=x+iy\): \[\frac{\bar z+2}{\bar z-1}=\frac{(x+2)-iy}{(x-1)-iy}\cdot\frac{(x-1)+iy}{(x-1)+iy}=\frac{(x^2+y^2+x-2)+3iy}{(x-1)^2+y^2} \] Real part is \(\displaystyle 4\): \[x^2+y^2+x-2=4\left[(x-1)^2+y^2\right] \] \[3x^2+3y^2-9x+6=0 \] \[\left(x-\tfrac32\right)^2+y^2=\tfrac14 \] NCERT_Solution_Class11_Maths_Exemplar_Ch5_Ex5-3_Q9 The point \(\displaystyle z=1\) makes the expression undefined, so it is removed.Answer: the locus is the circle with centre \(\displaystyle \left(\tfrac32,0\right)\) and radius \(\displaystyle \tfrac12\), the point \(\displaystyle z=1\) excluded.

  10. Exercise 10

    Show that the complex number z\displaystyle z, satisfying the condition arg⁡(z−1z+1)=π4\displaystyle \arg \left(\frac{z-1}{z+1}\right)=\frac{\pi}{4} lies on a circle.

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

    With \(\displaystyle z=x+iy\): \[\frac{z-1}{z+1}=\frac{(x-1)+iy}{(x+1)+iy}\cdot\frac{(x+1)-iy}{(x+1)-iy}=\frac{(x^2+y^2-1)+2iy}{(x+1)^2+y^2} \] \(\displaystyle \arg=\dfrac{\pi}{4}\) means real part \(\displaystyle =\) imaginary part \(\displaystyle >0\): \[x^2+y^2-1=2y>0 \] \[x^2+(y-1)^2=2,\quad y>0 \] NCERT_Solution_Class11_Maths_Exemplar_Ch5_Ex5-3_Q10 Answer: \(\displaystyle z\) lies on the circle with centre \(\displaystyle (0,1)\) and radius \(\displaystyle \sqrt2\), on the arc above the real axis.