NCERT Solutions · Class 11 Mathematics Complex Numbers and Quadratic Equations

Reduce the powers of \(\displaystyle i\) modulo $\displaystyle 4$, then cube.Since \(\displaystyle i^{4}=1\),\[i^{18}=i^{4(4)+2}=i^{2}=-1 \]For the second term, \(\displaystyle \dfrac{1}{i}=\dfrac{1}{i}\times\dfrac{i}{i}=\dfrac{i}{i^{2}}=-i\), so\[\left(\frac{1}{i}\right)^{25}=(-i)^{25}=(-1)^{25}i^{25}=-\,i^{4(6)+1}=-i \]Therefore the bracket is \(\displaystyle i^{18}+\left(\frac1i\right)^{25}=-1-i\), and\[(-1-i)^{3}=-(1+i)^{3} \]Now \(\displaystyle (1+i)^{2}=1+2i+i^{2}=2i\), so \(\displaystyle (1+i)^{3}=(1+i)(2i)=2i+2i^{2}=-2+2i\).\[(-1-i)^{3}=-(-2+2i)=2-2i \]Answer: \(\displaystyle 2-2i\).

Write both numbers in the form \(\displaystyle a+ib\) and multiply out.Let \(\displaystyle z_{1}=x_{1}+iy_{1}\) and \(\displaystyle z_{2}=x_{2}+iy_{2}\), so that\[\operatorname{Re}z_{1}=x_{1},\quad \operatorname{Im}z_{1}=y_{1},\quad \operatorname{Re}z_{2}=x_{2},\quad \operatorname{Im}z_{2}=y_{2} \]Multiplying, and using \(\displaystyle i^{2}=-1\):\[z_{1}z_{2}=(x_{1}+iy_{1})(x_{2}+iy_{2}) =x_{1}x_{2}+i\,x_{1}y_{2}+i\,y_{1}x_{2}+i^{2}y_{1}y_{2} \]\[=(x_{1}x_{2}-y_{1}y_{2})+i\,(x_{1}y_{2}+x_{2}y_{1}) \]This is already in the form \(\displaystyle a+ib\) with \(\displaystyle a,b\) real, so its real part is the bracket without \(\displaystyle i\):\[\operatorname{Re}(z_{1}z_{2})=x_{1}x_{2}-y_{1}y_{2} =\operatorname{Re}z_{1}\operatorname{Re}z_{2}-\operatorname{Im}z_{1}\operatorname{Im}z_{2} \]Hence \(\displaystyle \operatorname{Re}(z_{1}z_{2})=\operatorname{Re}z_{1}\operatorname{Re}z_{2}-\operatorname{Im}z_{1}\operatorname{Im}z_{2}\), as required.

Rationalise each fraction, subtract, then multiply.To put \(\displaystyle \dfrac{1}{a+ib}\) in standard form, multiply above and below by the conjugate \(\displaystyle a-ib\).\[\frac{1}{1-4i}=\frac{1}{1-4i}\times\frac{1+4i}{1+4i}=\frac{1+4i}{1^{2}+4^{2}}=\frac{1+4i}{17} \]\[\frac{2}{1+i}=\frac{2(1-i)}{1^{2}+1^{2}}=\frac{2(1-i)}{2}=1-i \]So the first bracket is\[\frac{1+4i}{17}-(1-i)=\frac{1+4i-17+17i}{17}=\frac{-16+21i}{17} \]The second factor:\[\frac{3-4i}{5+i}=\frac{(3-4i)(5-i)}{5^{2}+1^{2}}=\frac{15-3i-20i+4i^{2}}{26}=\frac{11-23i}{26} \]Multiply the two results:\[\frac{-16+21i}{17}\times\frac{11-23i}{26}=\frac{(-16+21i)(11-23i)}{442} \]\[(-16+21i)(11-23i)=-176+368i+231i-483i^{2}=(-176+483)+(368+231)i=307+599i \]\[=\frac{307+599i}{442} \]Since \(\displaystyle 442=2\times13\times17\) shares no factor with \(\displaystyle 307\) or \(\displaystyle 599\), the fraction is already in lowest terms.Answer: \(\displaystyle \dfrac{307}{442}+i\dfrac{599}{442}\).

Take conjugates and multiply — the product turns the two square roots into moduli.We are given\[x-iy=\sqrt{\frac{a-ib}{c-id}} \]Take the conjugate of both sides. Conjugation respects products, quotients and square roots, and \(\displaystyle \overline{x-iy}=x+iy\) since \(\displaystyle x,y\) are real:\[x+iy=\sqrt{\frac{a+ib}{c+id}} \]Multiply the two equations:\[(x-iy)(x+iy)=\sqrt{\frac{a-ib}{c-id}}\cdot\sqrt{\frac{a+ib}{c+id}} =\sqrt{\frac{(a-ib)(a+ib)}{(c-id)(c+id)}} \]The left side is \(\displaystyle x^{2}-i^{2}y^{2}=x^{2}+y^{2}\). On the right, \(\displaystyle (a-ib)(a+ib)=a^{2}+b^{2}\) and \(\displaystyle (c-id)(c+id)=c^{2}+d^{2}\). Hence\[x^{2}+y^{2}=\sqrt{\frac{a^{2}+b^{2}}{c^{2}+d^{2}}} \]Squaring both sides:\[\left(x^{2}+y^{2}\right)^{2}=\frac{a^{2}+b^{2}}{c^{2}+d^{2}} \]Hence \(\displaystyle \left(x^{2}+y^{2}\right)^{2}=\dfrac{a^{2}+b^{2}}{c^{2}+d^{2}}\), as required.

Simplify, then take the modulus.Substitute \(\displaystyle z_1 = 2 - i \) and \(\displaystyle z_2 = 1 + i \) into the numerator and the denominator separately.Numerator: \[z_1 + z_2 + 1 = (2 - i) + (1 + i) + 1 = 4 + 0i = 4 \]Denominator: \[z_1 - z_2 + 1 = (2 - i) - (1 + i) + 1 = 2 - 2i \]So the quotient is \(\displaystyle \dfrac{4}{2 - 2i} \).For complex numbers, \(\displaystyle \left|\dfrac{z}{w}\right| = \dfrac{|z|}{|w|} \), and \(\displaystyle |x + iy| = \sqrt{x^2 + y^2} \). Hence \[\left|\frac{4}{2 - 2i}\right| = \frac{|4|}{|2 - 2i|} = \frac{4}{\sqrt{2^2 + (-2)^2}} = \frac{4}{\sqrt{8}} = \frac{4}{2\sqrt{2}} = \sqrt{2} \]The required modulus is \(\displaystyle \sqrt{2} \).

Modulus method.For any complex number \(\displaystyle z = a + ib \) we have \(\displaystyle |z|^2 = a^2 + b^2 \). So instead of separating real and imaginary parts, take the modulus of both sides and square.Use the two rules \(\displaystyle \left|\dfrac{z}{w}\right| = \dfrac{|z|}{|w|} \) and \(\displaystyle |z^n| = |z|^n \). Since \(\displaystyle x \) is real, \(\displaystyle 2x^2 + 1 \) is a positive real number, so its modulus is itself.\[|a + ib| = \left|\frac{(x+i)^2}{2x^2+1}\right| = \frac{|x+i|^2}{2x^2+1} \]Now \(\displaystyle |x + i| = \sqrt{x^2 + 1} \), so \(\displaystyle |x + i|^2 = x^2 + 1 \), giving \[|a + ib| = \frac{x^2+1}{2x^2+1} \]Squaring both sides: \[a^2 + b^2 = |a+ib|^2 = \frac{\left(x^2+1\right)^2}{\left(2x^2+1\right)^2} \]Hence \(\displaystyle a^2 + b^2 = \dfrac{\left(x^{2}+1\right)^{2}}{\left(2 x^{2}+1\right)^{2}} \), as required.

Reduce each expression to \(\displaystyle a + ib \) form.Here \(\displaystyle z_1 = 2 - i \), so \(\displaystyle \bar{z}_1 = 2 + i \), and \(\displaystyle z_2 = -2 + i \).(i) First the product: \[z_1 z_2 = (2-i)(-2+i) = -4 + 2i + 2i - i^2 = -4 + 4i + 1 = -3 + 4i \]Now divide by \(\displaystyle \bar{z}_1 = 2 + i \). To make the denominator real, multiply numerator and denominator by the conjugate \(\displaystyle 2 - i \): \[\frac{-3+4i}{2+i} = \frac{(-3+4i)(2-i)}{(2+i)(2-i)} = \frac{-6 + 3i + 8i - 4i^2}{4 + 1} = \frac{-2 + 11i}{5} = -\frac{2}{5} + \frac{11}{5}i \]The real part is the coefficient standing free of \(\displaystyle i \).\(\displaystyle \operatorname{Re}\!\left(\dfrac{z_1 z_2}{\bar{z}_1}\right) = -\dfrac{2}{5} \).(ii) For any complex number, \(\displaystyle z\bar{z} = |z|^2 \), which is real. Here \[z_1 \bar{z}_1 = (2-i)(2+i) = 4 - i^2 = 5 \] so \[\frac{1}{z_1 \bar{z}_1} = \frac{1}{5} = \frac{1}{5} + 0\,i \]There is no imaginary part.\(\displaystyle \operatorname{Im}\!\left(\dfrac{1}{z_1 \bar{z}_1}\right) = 0 \).

Equate real and imaginary parts.The conjugate of \(\displaystyle -6 - 24i \) is \(\displaystyle -6 + 24i \). So the condition is \[(x - iy)(3 + 5i) = -6 + 24i \]Expand the left side, using \(\displaystyle i^2 = -1 \): \[(x - iy)(3+5i) = 3x + 5xi - 3iy - 5i^2 y = (3x + 5y) + i(5x - 3y) \]Two complex numbers are equal only when their real parts are equal and their imaginary parts are equal. Since \(\displaystyle x, y \) are real: \[3x + 5y = -6 \qquad \text{and} \qquad 5x - 3y = 24 \]Multiply the first by \(\displaystyle 3 \) and the second by \(\displaystyle 5 \), then add to eliminate \(\displaystyle y \): \[9x + 15y = -18, \qquad 25x - 15y = 120 \] \[34x = 102 \;\Rightarrow\; x = 3 \]Substituting \(\displaystyle x = 3 \) into \(\displaystyle 3x + 5y = -6 \): \[9 + 5y = -6 \;\Rightarrow\; 5y = -15 \;\Rightarrow\; y = -3 \]Check: \(\displaystyle (3 + 3i)(3 + 5i) = 9 + 15i + 9i + 15i^2 = -6 + 24i \), which is indeed the conjugate of \(\displaystyle -6 - 24i \).\(\displaystyle x = 3 \) and \(\displaystyle y = -3 \).

Combine over a common denominator, then take the modulus.The common denominator is \(\displaystyle (1-i)(1+i) = 1 - i^2 = 2 \), so \[\frac{1+i}{1-i} - \frac{1-i}{1+i} = \frac{(1+i)^2 - (1-i)^2}{(1-i)(1+i)} \]Now \(\displaystyle (1+i)^2 = 1 + 2i + i^2 = 2i \) and \(\displaystyle (1-i)^2 = 1 - 2i + i^2 = -2i \). Hence the numerator is \(\displaystyle 2i - (-2i) = 4i \), and \[\frac{4i}{2} = 2i \]Finally \(\displaystyle |2i| = \sqrt{0^2 + 2^2} = 2 \).The modulus is \(\displaystyle 2 \).

Expand the cube and compare parts.Using \(\displaystyle (p+q)^3 = p^3 + 3p^2q + 3pq^2 + q^3 \) with \(\displaystyle p = x \), \(\displaystyle q = iy \), and \(\displaystyle i^2 = -1 \), \(\displaystyle i^3 = -i \): \[(x+iy)^3 = x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3 = x^3 + 3ix^2 y - 3xy^2 - iy^3 \] \[= \left(x^3 - 3xy^2\right) + i\left(3x^2 y - y^3\right) \]Since \(\displaystyle x, y, u, v \) are real and \(\displaystyle (x+iy)^3 = u + iv \), equating real and imaginary parts gives \[u = x^3 - 3xy^2 = x\left(x^2 - 3y^2\right), \qquad v = 3x^2 y - y^3 = y\left(3x^2 - y^2\right) \]Therefore (with \(\displaystyle x \neq 0 \), \(\displaystyle y \neq 0 \) so the quotients are defined) \[\frac{u}{x} + \frac{v}{y} = \left(x^2 - 3y^2\right) + \left(3x^2 - y^2\right) = 4x^2 - 4y^2 = 4\left(x^2 - y^2\right) \]Hence \(\displaystyle \dfrac{u}{x} + \dfrac{v}{y} = 4\left(x^{2}-y^{2}\right) \), as required.