SolveItClass 11 · NCERT

NCERT Solutions · Class 11 Mathematics Complex Numbers and Quadratic Equations

28 questions · 28 still being checked

EXERCISE 4.1 11–14 (part 2 of 4)

  1. Find the multiplicative inverse of each of the complex numbers given in the Exercises $\displaystyle 11$ to 13.

    Exercise 11

    4−3i\displaystyle 4-3 i

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    NCERT’s answer
    $\displaystyle \frac{4}{25}+i \frac{3}{25}$
    Multiplicative inverse rule. For \(\displaystyle z=a+ib\neq 0\),\[z^{-1}=\frac{\bar z}{|z|^{2}}=\frac{a-ib}{a^{2}+b^{2}} \]Here \(\displaystyle z=4-3i\), so \(\displaystyle \bar z=4+3i\) and \(\displaystyle |z|^{2}=4^{2}+(-3)^{2}=16+9=25\).\[z^{-1}=\frac{4+3i}{25}=\frac{4}{25}+i\frac{3}{25} \]Check: \(\displaystyle (4-3i)\left(\frac{4}{25}+\frac{3}{25}i\right)=\frac{16+12i-12i-9i^{2}}{25}=\frac{25}{25}=1\).Answer: \(\displaystyle \dfrac{4}{25}+i\dfrac{3}{25}\).

  2. Exercise 12

    5+3i\displaystyle \sqrt{5}+3 i

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    NCERT’s answer
    $\displaystyle \frac{\sqrt{5}}{14}-i \frac{3}{14}$
    Multiplicative inverse rule. For \(\displaystyle z=a+ib\neq 0\), \(\displaystyle z^{-1}=\dfrac{\bar z}{|z|^{2}}=\dfrac{a-ib}{a^{2}+b^{2}}\).Here \(\displaystyle z=\sqrt{5}+3i\), so \(\displaystyle \bar z=\sqrt{5}-3i\) and\[|z|^{2}=\left(\sqrt{5}\right)^{2}+3^{2}=5+9=14 \]Hence\[z^{-1}=\frac{\sqrt{5}-3i}{14}=\frac{\sqrt{5}}{14}-i\frac{3}{14} \]Answer: \(\displaystyle \dfrac{\sqrt{5}}{14}-i\dfrac{3}{14}\).

  3. Exercise 13

    −i\displaystyle -i

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    NCERT’s answer
    $\displaystyle 0+i 1$
    Multiplicative inverse rule. For \(\displaystyle z=a+ib\neq 0\), \(\displaystyle z^{-1}=\dfrac{\bar z}{|z|^{2}}\).Here \(\displaystyle z=-i=0+i(-1)\), so \(\displaystyle \bar z=i\) and \(\displaystyle |z|^{2}=0^{2}+(-1)^{2}=1\).\[z^{-1}=\frac{i}{1}=i \]Check: \(\displaystyle (-i)(i)=-i^{2}=1\).Answer: \(\displaystyle 0+i\), i.e. \(\displaystyle i\).

  4. Exercise 14

    Express the following expression in the form of a+ib\displaystyle a+i b : (3+i5)(3−i5)(3+2i)−(3−i2)\frac{(3+i \sqrt{5})(3-i \sqrt{5})}{(\sqrt{3}+\sqrt{2} i)-(\sqrt{3}-i \sqrt{2})}

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    NCERT’s answer
    $\displaystyle 0-i \frac{7 \sqrt{2}}{2}$
    Simplify numerator and denominator separately, then rationalise.Numerator — a difference of squares, using \(\displaystyle i^{2}=-1\):\[(3+i\sqrt{5})(3-i\sqrt{5})=3^{2}-\left(i\sqrt{5}\right)^{2}=9-5i^{2}=9+5=14 \]Denominator:\[(\sqrt{3}+\sqrt{2}\,i)-(\sqrt{3}-i\sqrt{2})=\sqrt{3}-\sqrt{3}+\sqrt{2}\,i+\sqrt{2}\,i=2\sqrt{2}\,i \]So the expression is \(\displaystyle \dfrac{14}{2\sqrt{2}\,i}=\dfrac{7}{\sqrt{2}\,i}\). Multiply numerator and denominator by \(\displaystyle i\) to clear \(\displaystyle i\) from the denominator:\[\frac{7}{\sqrt{2}\,i}\times\frac{i}{i}=\frac{7i}{\sqrt{2}\,i^{2}}=-\frac{7i}{\sqrt{2}}=-\frac{7\sqrt{2}}{2}\,i \]Answer: \(\displaystyle 0-i\dfrac{7\sqrt{2}}{2}\) (that is, \(\displaystyle -\dfrac{7}{\sqrt{2}}i\)).