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NCERT Solutions · Class 11 Mathematics Complex Numbers and Quadratic Equations

28 exercises · 28 still being checked

EXERCISE 4.1 1–10 (part 1 of 4)

  1. Express each of the complex number given in the Exercises $\displaystyle 1$ to $\displaystyle 10$ in the form \(\displaystyle a+i b\).

    Exercise 1

    \(\displaystyle (5 i)\left(-\frac{3}{5} i\right)\)

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    NCERT’s answer
    $\displaystyle 3+i 0$
    Multiply, then use \(\displaystyle i^{2}=-1\).\[(5i)\left(-\frac{3}{5}i\right)=5\times\left(-\frac{3}{5}\right)\times i^{2}=-3\,i^{2}=-3(-1)=3 \]The imaginary part is \(\displaystyle 0\), so in the form \(\displaystyle a+ib\) we write \(\displaystyle a=3,\ b=0\).Answer: \(\displaystyle 3+i0\).

  2. Exercise 2

    \(\displaystyle i^{9}+i^{19}\)

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    NCERT’s answer
    $\displaystyle 0+i 0$
    Reduce each power of \(\displaystyle i\) modulo 4.Since \(\displaystyle i^{4}=1\), any power of \(\displaystyle i\) repeats with period $\displaystyle 4$: write the exponent as \(\displaystyle 4q+r\) and keep only \(\displaystyle i^{r}\).\[i^{9}=i^{4(2)+1}=\left(i^{4}\right)^{2}\cdot i=i, \qquad i^{19}=i^{4(4)+3}=\left(i^{4}\right)^{4}\cdot i^{3}=i^{3}=-i \]Hence\[i^{9}+i^{19}=i+(-i)=0 \]Answer: \(\displaystyle 0+i0\).

  3. Exercise 3

    \(\displaystyle i^{-39}\)

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    NCERT’s answer
    $\displaystyle 0+i 1$
    Turn the negative power into a positive one using \(\displaystyle i^{4}=1\).Multiplying by \(\displaystyle i^{40}=\left(i^{4}\right)^{10}=1\) changes nothing:\[i^{-39}=i^{-39}\cdot i^{40}=i^{-39+40}=i^{1}=i \](Equivalently, \(\displaystyle i^{-39}=\dfrac{1}{i^{39}}=\dfrac{1}{i^{3}}=\dfrac{1}{-i}=\dfrac{1}{-i}\times\dfrac{i}{i}=\dfrac{i}{-i^{2}}=i\).)Answer: \(\displaystyle 0+i\).

  4. Exercise 4

    \(\displaystyle 3(7+i 7)+i(7+i 7)\)

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    NCERT’s answer
    $\displaystyle 14+28 i$
    Expand both products and collect real and imaginary parts.\[3(7+i7)+i(7+i7)=21+21i+7i+7i^{2} \]Using \(\displaystyle i^{2}=-1\), the term \(\displaystyle 7i^{2}=-7\):\[=(21-7)+(21+7)i=14+28i \]Answer: \(\displaystyle 14+i28\).

  5. Exercise 5

    \(\displaystyle (1-i)-(-1+i 6)\)

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    NCERT’s answer
    $\displaystyle 2-7 i$
    Subtract real parts from real parts and imaginary parts from imaginary parts.\[(1-i)-(-1+i6)=1-i+1-6i=(1+1)+(-1-6)i=2-7i \]Answer: \(\displaystyle 2-i7\).

  6. Exercise 6

    \(\displaystyle \left(\frac{1}{5}+i \frac{2}{5}\right)-\left(4+i \frac{5}{2}\right)\)

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    NCERT’s answer
    $\displaystyle -\frac{19}{5}-\frac{21 i}{10}$
    Subtract componentwise.\[\left(\frac{1}{5}+i\frac{2}{5}\right)-\left(4+i\frac{5}{2}\right) =\left(\frac{1}{5}-4\right)+i\left(\frac{2}{5}-\frac{5}{2}\right) \]Real part: \(\displaystyle \dfrac{1}{5}-4=\dfrac{1-20}{5}=-\dfrac{19}{5}\).Imaginary part: \(\displaystyle \dfrac{2}{5}-\dfrac{5}{2}=\dfrac{4-25}{10}=-\dfrac{21}{10}\).Answer: \(\displaystyle -\dfrac{19}{5}-i\dfrac{21}{10}\).

  7. Exercise 7

    \(\displaystyle \left[\left(\frac{1}{3}+i \frac{7}{3}\right)+\left(4+i \frac{1}{3}\right)\right]-\left(-\frac{4}{3}+i\right)\)

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    NCERT’s answer
    $\displaystyle \frac{17}{3}+i \frac{5}{3}$
    Add inside the bracket first, then subtract componentwise.\[\left(\frac{1}{3}+i\frac{7}{3}\right)+\left(4+i\frac{1}{3}\right) =\left(\frac{1}{3}+4\right)+i\left(\frac{7}{3}+\frac{1}{3}\right) =\frac{13}{3}+i\frac{8}{3} \]Now subtract \(\displaystyle -\dfrac{4}{3}+i\):\[\left(\frac{13}{3}+i\frac{8}{3}\right)-\left(-\frac{4}{3}+i\right) =\left(\frac{13}{3}+\frac{4}{3}\right)+i\left(\frac{8}{3}-1\right) =\frac{17}{3}+i\frac{5}{3} \]Answer: \(\displaystyle \dfrac{17}{3}+i\dfrac{5}{3}\).

  8. Exercise 8

    \(\displaystyle (1-i)^{4}\)

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    NCERT’s answer
    $\displaystyle -4+i 0$
    Square once, then square again — it is shorter than a full binomial expansion.\[(1-i)^{2}=1-2i+i^{2}=1-2i-1=-2i \]Therefore\[(1-i)^{4}=\left[(1-i)^{2}\right]^{2}=(-2i)^{2}=4i^{2}=-4 \]Answer: \(\displaystyle -4+i0\).

  9. Exercise 9

    \(\displaystyle \left(\frac{1}{3}+3 i\right)^{3}\)

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    NCERT’s answer
    $\displaystyle -\frac{242}{27}-26 i$
    Use the identity \(\displaystyle (a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}\) with \(\displaystyle a=\dfrac{1}{3}\) and \(\displaystyle b=3i\).\[a^{3}=\frac{1}{27},\qquad 3a^{2}b=3\cdot\frac{1}{9}\cdot 3i=i,\qquad 3ab^{2}=3\cdot\frac{1}{3}\cdot 9i^{2}=-9,\qquad b^{3}=27i^{3}=-27i \]Adding, and grouping real with real:\[\left(\frac{1}{3}+3i\right)^{3}=\left(\frac{1}{27}-9\right)+(1-27)i =\frac{1-243}{27}-26i=-\frac{242}{27}-26i \]Answer: \(\displaystyle -\dfrac{242}{27}-i26\).

  10. Exercise 10

    \(\displaystyle \left(-2-\frac{1}{3} i\right)^{3}\)

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    NCERT’s answer
    $\displaystyle \frac{-22}{3}-i \frac{107}{27}$
    Pull out the minus sign first, then cube with \(\displaystyle (a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}\).\[\left(-2-\frac{1}{3}i\right)^{3}=-\left(2+\frac{1}{3}i\right)^{3} \]With \(\displaystyle a=2\) and \(\displaystyle b=\dfrac{i}{3}\):\[a^{3}=8,\qquad 3a^{2}b=3\cdot 4\cdot\frac{i}{3}=4i,\qquad 3ab^{2}=3\cdot 2\cdot\frac{i^{2}}{9}=-\frac{2}{3},\qquad b^{3}=\frac{i^{3}}{27}=-\frac{i}{27} \]\[\left(2+\frac{1}{3}i\right)^{3}=\left(8-\frac{2}{3}\right)+i\left(4-\frac{1}{27}\right) =\frac{22}{3}+i\frac{107}{27} \]Restoring the minus sign:\[\left(-2-\frac{1}{3}i\right)^{3}=-\frac{22}{3}-i\frac{107}{27} \]Answer: \(\displaystyle -\dfrac{22}{3}-i\dfrac{107}{27}\).