NCERT Solutions: Class 11 Maths Chapter 4 - Complex Numbers and Quadratic Equations

$(5i)\left(-\dfrac35 i\right)=-3i^2=3$.

$i^9=i^{8}i=i$ and $i^{19}=i^{16}i^3=-i$. Therefore $i^9+i^{19}=0$.

Since powers of $i$ repeat every 4 and $-39\equiv1\pmod4$, $i^{-39}=i$.

$3(7+7i)+i(7+7i)=21+21i+7i+7i^2=21+28i-7=14+28i$.

$(1-i)-(-1+6i)=1-i+1-6i=2-7i$.

Subtract real and imaginary parts separately: $\dfrac15-4=-\dfrac{19}{5}$ and $\dfrac25-\dfrac52=\dfrac4{10}-\dfrac{25}{10}=-\dfrac{21}{10}$.

The bracket equals $\dfrac{13}{3}+\dfrac{8}{3}i$. Subtracting $-\dfrac43+i$ gives real part $\dfrac{13}{3}+\dfrac43=\dfrac{17}{3}$ and imaginary part $\dfrac83-1=\dfrac53$.

$(1-i)^2=1-2i+i^2=-2i$. Hence $(1-i)^4=(-2i)^2=4i^2=-4$.

Using $(a+ib)^3=(a^3-3ab^2)+i(3a^2b-b^3)$ with $a=\dfrac13$, $b=3$, the real part is $\dfrac1{27}-9=-\dfrac{242}{27}$ and the imaginary part is $1-27=-26$.

With $a=-2$ and $b=-\dfrac13$, $(a+ib)^3=(a^3-3ab^2)+i(3a^2b-b^3)$. The real part is $-8+\dfrac23=-\dfrac{22}{3}$ and the imaginary part is $-4+\dfrac1{27}=-\dfrac{107}{27}$.

$(4-3i)^{-1}=\dfrac{4+3i}{4^2+3^2}=\dfrac{4+3i}{25}$.

$(\sqrt5+3i)^{-1}=\dfrac{\sqrt5-3i}{(\sqrt5)^2+3^2}=\dfrac{\sqrt5-3i}{14}$.

$(-i)i=-i^2=1$, so the multiplicative inverse of $-i$ is $i$.

The numerator is $3^2+(\sqrt5)^2=14$. The denominator is $2\sqrt2 i$. Thus the expression is $\dfrac{14}{2\sqrt2 i}=\dfrac{7}{\sqrt2 i}=-\dfrac{7}{\sqrt2}i=-\dfrac{7\sqrt2}{2}i$.

$i^{18}=i^2=-1$ and $\dfrac1i=-i$, so $\left(\dfrac1i\right)^{25}=(-i)^{25}=-i$. Therefore the expression is $(-1-i)^3=2-2i$.

Let $z_1=a+ib$ and $z_2=c+id$. Then $z_1z_2=(ac-bd)+i(ad+bc)$. Hence $\operatorname{Re}(z_1z_2)=ac-bd=(\operatorname{Re}z_1)(\operatorname{Re}z_2)-(\operatorname{Im}z_1)(\operatorname{Im}z_2)$.

Rationalising each denominator and multiplying gives $\left(\dfrac{1+4i}{17}-(1-i)\right)\left(\dfrac{(3-4i)(5-i)}{26}\right)=\left(-\dfrac{16}{17}+\dfrac{21}{17}i\right)\left(\dfrac{11}{26}-\dfrac{23}{26}i\right)=\dfrac{307}{442}+\dfrac{599}{442}i$.

Squaring, $(x-iy)^2=\dfrac{a-ib}{c-id}$. Now take modulus squared on both sides. Then $|(x-iy)^2|^2=|x-iy|^4=(x^2+y^2)^2$, while $\left|\dfrac{a-ib}{c-id}\right|^2=\dfrac{|a-ib|^2}{|c-id|^2}=\dfrac{a^2+b^2}{c^2+d^2}$. Hence $(x^2+y^2)^2=\dfrac{a^2+b^2}{c^2+d^2}$.

$z_1+z_2+1=(2-i)+(1+i)+1=4$ and $z_1-z_2+1=(2-i)-(1+i)+1=2-2i$. Thus the quotient is $\dfrac{4}{2-2i}=1+i$, whose modulus is $\sqrt{1^2+1^2}=\sqrt2$.

$a^2+b^2=|a+ib|^2=\left|\dfrac{(x+i)^2}{2x^2+1}\right|^2=\dfrac{|x+i|^4}{(2x^2+1)^2}=\dfrac{(x^2+1)^2}{(2x^2+1)^2}$.

$z_1z_2=(2-i)(-2+i)=-3+4i$ and $\overline{z_1}=2+i$. Hence $\dfrac{z_1z_2}{\overline{z_1}}=\dfrac{-3+4i}{2+i}=-\dfrac25+\dfrac{11}{5}i$, so the real part is $-\dfrac25$. Also $z_1\overline{z_1}=|z_1|^2=5$, so $\dfrac1{z_1\overline{z_1}}=\dfrac15$ has imaginary part 0.

The conjugate of $-6-24i$ is $-6+24i$. Now $(x-iy)(3+5i)=(3x+5y)+i(5x-3y)$. Equating real and imaginary parts gives $3x+5y=-6$ and $5x-3y=24$. Solving, $x=3$ and $y=-3$.

$\dfrac{1+i}{1-i}=i$ and $\dfrac{1-i}{1+i}=-i$. Their difference is $2i$, whose modulus is 2.

$(x+iy)^3=(x^3-3xy^2)+i(3x^2y-y^3)$, so $u=x^3-3xy^2$ and $v=3x^2y-y^3$. Therefore $\dfrac ux+\dfrac vy=(x^2-3y^2)+(3x^2-y^2)=4x^2-4y^2=4(x^2-y^2)$.

Since $|\beta|=1$, $\overline{\beta}=\dfrac1\beta$. Now $1-\overline{\alpha}\beta=\beta(\overline{\beta}-\overline{\alpha})$. Hence $|1-\overline{\alpha}\beta|=|\beta|\,|\overline{\beta}-\overline{\alpha}|=|\beta-\alpha|$. Therefore the required modulus is 1.

$|1-i|=\sqrt2$, so the equation becomes $(\sqrt2)^x=2^x$, i.e. $2^{x/2}=2^x$. Hence $x/2=x$, giving $x=0$. Since the question asks for non-zero integral solutions, there are none.

Taking modulus squared on both sides, $|(a+ib)(c+id)(e+if)(g+ih)|^2=|A+iB|^2$. Since modulus is multiplicative, the left side is $(a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2)$ and the right side is $A^2+B^2$.

$\dfrac{1+i}{1-i}=i$. Therefore the equation becomes $i^m=1$. The least positive integer $m$ for which this holds is $m=4$.