Samacheer Class 12 Maths - Complex Numbers

(4) $i^{59}=i^{4(14)+3}=-i$, so $1/i^{59}=1/(-i)=i$. Their sum is $0$.

(5) $i\,i^2\cdots i^{2000}=i^{1+2+\cdots+2000}=i^{2000(2001)/2}=i^{2001000}=1$, since $2001000$ is divisible by $4$.

(6) The powers from $i^{51}$ through $i^{58}$ form two complete cycles and sum to zero. Thus the sum is $i^{59}+i^{60}=-i+1=1-i$.

The first number is $(3x-2y+5)+i(-x+y+2)$, and the second is $(2x-y+3)+i(2y+2)$.

Equating real parts gives $3x-2y+5=2x-y+3$, hence $x-y=-2$.

Equating imaginary parts gives $-x+y+2=2y+2$, hence $x+y=0$. Solving, $x=-1$ and $y=1$.

Check: both original expressions become $0+5i$.

(i) $(z_1+z_2)+z_3=(1-7i)+5=6-7i$, while $z_1+(z_2+z_3)=(1-3i)+(5-4i)=6-7i$.

(ii) $(z_1z_2)z_3=[(1-3i)(-4i)]5=(-12-4i)5=-60-20i$. Also $z_1(z_2z_3)=(1-3i)(-20i)=-60-20i$.

(i) $z_1(z_2+z_3)=3(5-3i)=15-9i$. Also $z_1z_2+z_1z_3=-21i+(15+12i)=15-9i$.

(ii) $(z_1+z_2)z_3=(3-7i)(5+4i)=43-23i$. Also $z_1z_3+z_2z_3=(15+12i)+(28-35i)=43-23i$.

For $z=a+bi$, the additive inverse is $-z$, and the multiplicative inverse is $z^{-1}=\overline z/|z|^2$.

$z_1$: additive inverse $=-2-5i$; multiplicative inverse $=(2-5i)/(2^2+5^2)=(2-5i)/29$.

$z_2$: additive inverse $=3+4i$; multiplicative inverse $=(-3+4i)/(3^2+4^2)=(-3+4i)/25$.

$z_3$: additive inverse $=-1-i$; multiplicative inverse $=(1-i)/(1^2+1^2)=(1-i)/2$.

(i) For $z\ne0$, $1/z=(x-iy)/(x^2+y^2)$, so $\operatorname{Re}(1/z)=x/(x^2+y^2)$.

(ii) $i\overline z=i(x-iy)=y+ix$, hence $\operatorname{Re}(i\overline z)=y$.

(iii) $3z+4\overline z-4i=3(x+iy)+4(x-iy)-4i=7x-i(y+4)$. Its imaginary part is $-y-4$.

$z_1z_2=(2-i)(-4+3i)=-5+10i$. Therefore $(z_1z_2)^{-1}=(-5-10i)/125=(-1-2i)/25$.

$z_1/z_2=(2-i)/(-4+3i)=(-11-2i)/25$. Its inverse is $z_2/z_1=(-4+3i)/(2-i)=(-11+2i)/5$.

Verification: $(-5+10i)(-1-2i)/25=1$ and $[(-11-2i)/25][(-11+2i)/5]=1$.

(i) Let $A=(2+i\sqrt3)^{10}-(2-i\sqrt3)^{10}$. Then $\overline A=(2-i\sqrt3)^{10}-(2+i\sqrt3)^{10}=-A$. Therefore $A$ is purely imaginary.

(ii) $\dfrac{19-7i}{9+i}=2-i$ and $\dfrac{20-5i}{7-6i}=2+i$. The required expression is $(2-i)^{12}+(2+i)^{12}$. Its two terms are conjugates, so their sum equals twice a real part and is real.

(i) $\left|\dfrac{2i}{3+4i}\right|=2/5$.

(ii) $\dfrac{2-i}{1+i}=(1-3i)/2$ and $\dfrac{1-2i}{1-i}=(3-i)/2$, so the sum is $2-2i$ and its modulus is $2\sqrt2$.

(iii) $|(1-i)^{10}|=|1-i|^{10}=(\sqrt2)^{10}=32$.

(iv) $|2i(3-4i)(4-3i)|=2\cdot5\cdot5=50$.

Let $w=6-8i$. Then $|w|=10$.

By the reverse triangle inequality, $|z+w|\ge||w|-|z||=|10-3|=7$.

By the triangle inequality, $|z+w|\le|z|+|w|=3+10=13$.

$|z^2|=|z|^2=1$. By the reverse triangle inequality, $|z^2-3|\ge|3-|z^2||=2$.

By the triangle inequality, $|z^2-3|\le|z^2|+3=4$.

Let $w=6+8i$, so $|w|=10$.

The reverse triangle inequality gives $|z+w|\ge||w|-|z||=8$, and the triangle inequality gives $|z+w|\le|z|+|w|=12$.

From the moduli, $\overline{z_1}=1/z_1$, $\overline{z_2}=4/z_2$, and $\overline{z_3}=9/z_3$.

$z_1z_2z_3(\overline{z_1}+\overline{z_2}+\overline{z_3})=z_2z_3+4z_1z_3+9z_1z_2$.

Taking moduli gives the required modulus as $|z_1z_2z_3|\,|\overline{z_1+z_2+z_3}|=(1)(2)(3)(1)=6$.

Let z = x+iy. The three vertices are A=z, B=i z, C=(1+i)z. The area equals (1/2)|Im[(B-A)·conj(C-A)]|. Compute B-A = z(i-1), C-A = iz. Then (B-A)·conj(C-A)=z(i-1)·conj(iz)=z(i-1)·(-i)conj(z) = -i|z|^2(i-1) = |z|^2(1+i). Its imaginary part = |z|^2. Hence area = (1/2)|z|^2 = 50, so |z|^2 = 100 and |z| = 10.

$z=0$ is one solution. For $z\ne0$, write $z=r\operatorname{cis}\theta$ with $r>0$.

$r^3\operatorname{cis}(3\theta)=-2r\operatorname{cis}(-\theta)=2r\operatorname{cis}(\pi-\theta)$. Hence $r^2=2$ and $4\theta=(2k+1)\pi$.

Thus $r=\sqrt2$ and $\theta=(2k+1)\pi/4$ for $k=0,1,2,3$, giving four distinct nonzero solutions. Together with $z=0$, there are five.

The condition is $|z-4i|=|z+4i|$.

Squaring, $x^2+(y-4)^2=x^2+(y+4)^2$. This simplifies to $-8y=8y$, so $y=0$.

$2z+1=(2x+1)+2iy$ and $iz+1=(1-y)+ix$.

For $(a+ib)/(c+id)$, the imaginary numerator after rationalising is $bc-ad$. Here it is $2y(1-y)-x(2x+1)$.

Setting it to zero gives $2y-2y^2-2x^2-x=0$, or $2x^2+2y^2+x-2y=0$.

(i) $iz=-y+ix$, so $[\operatorname{Re}(iz)]^2=y^2=3$.

(ii) $(1-i)z+1=(x+y+1)+i(y-x)$, so $y-x=0$.

(iii) $x^2+(y+1)^2=(x-1)^2+y^2$, which reduces to $x+y=0$.

(iv) $z^{-1}=\overline z/|z|^2$. Since $z\ne0$, $\overline z=z^{-1}$ implies $|z|^2=1$, hence $x^2+y^2=1$.

(i) $|z-(2+i)|=3$: centre $2+i$, radius $3$.

(ii) $2|z+1-2i|=2$, so $|z-(-1+2i)|=1$: centre $-1+2i$, radius $1$.

(iii) $3|z-2+4i|=8$, so $|z-(2-4i)|=8/3$: centre $2-4i$, radius $8/3$.

(i) $|x-4+iy|=16$, so $(x-4)^2+y^2=256$.

(ii) $[(x-4)^2+y^2]-[(x-1)^2+y^2]=16$. Simplifying gives $-6x+15=16$, hence $x=-1/6$.

(i) Modulus $4$ and argument $\pi/3$, so $4\operatorname{cis}(\pi/3)$.

(ii) Modulus $2\sqrt3$ and argument $-\pi/6$, so $2\sqrt3\operatorname{cis}(-\pi/6)$.

(iii) Modulus $2\sqrt2$ and principal argument $-3\pi/4$, so $2\sqrt2\operatorname{cis}(-3\pi/4)$.

(iv) $i-1=\sqrt2\operatorname{cis}(3\pi/4)$. Dividing by $\operatorname{cis}(\pi/3)$ gives $\sqrt2\operatorname{cis}(5\pi/12)$.

(i) By multiplication of polar forms, the product is $\operatorname{cis}(\pi/6+\pi/12)=\operatorname{cis}(\pi/4)=(1+i)/\sqrt2$.

(ii) The quotient is $\frac12\operatorname{cis}(-\pi/6-\pi/3)=\frac12\operatorname{cis}(-\pi/2)=-i/2$.

Taking moduli of both sides gives $\prod_{r=1}^n\sqrt{x_r^2+y_r^2}=\sqrt{a^2+b^2}$. Squaring proves (i).

Taking arguments of both sides gives $\sum_{r=1}^n\arg(x_r+iy_r)=\arg(a+ib)+2k\pi$. Using $\arg(x+iy)=\tan^{-1}(y/x)$ with the appropriate quadrant gives (ii).

Write $\operatorname{cis}(2\theta)=e^{2i\theta}$. Then $1+z=(1-z)e^{2i\theta}$, so $z(1+e^{2i\theta})=e^{2i\theta}-1$.

$e^{2i\theta}-1=2ie^{i\theta}\sin\theta$ and $1+e^{2i\theta}=2e^{i\theta}\cos\theta$. Therefore $z=i\tan\theta$.

Let $u=e^{i\alpha}$, $v=e^{i\beta}$, and $w=e^{i\gamma}$. The hypotheses give $u+v+w=0$.

Using $u^3+v^3+w^3-3uvw=(u+v+w)(u^2+v^2+w^2-uv-vw-wu)$, we get $u^3+v^3+w^3=3uvw$.

Thus $e^{3i\alpha}+e^{3i\beta}+e^{3i\gamma}=3e^{i(\alpha+\beta+\gamma)}$. Equating real and imaginary parts proves (i) and (ii).

Rationalising, $\dfrac{z-i}{z+2}=\dfrac{[x+i(y-1)][x+2-iy]}{(x+2)^2+y^2}$.

Its real numerator is $x^2+2x+y^2-y$, and its imaginary numerator is $2y-x-2$.

Argument $\pi/4$ means imaginary part equals real part. Hence $2y-x-2=x^2+2x+y^2-y$, which rearranges to $x^2+y^2+3x-3y+2=0$.

Let $A=a+b\omega+c\omega^2$. Since $\omega^3=1$, $\omega^2A=b+c\omega+a\omega^2$ and $\omega A=c+a\omega+b\omega^2$.

Therefore the left side is $A/(\omega^2A)+A/(\omega A)=1/\omega^2+1/\omega=\omega+\omega^2=-1$, using $1+\omega+\omega^2=0$.

The two bases are $\operatorname{cis}(\pi/6)$ and $\operatorname{cis}(-\pi/6)$.

By De Moivre's theorem, the sum is $\operatorname{cis}(5\pi/6)+\operatorname{cis}(-5\pi/6)=2\cos(5\pi/6)=-\sqrt3$.

Let $a=\pi/10$. Since $\cos a/(1+\sin a)=\tan(\pi/4-a/2)=\tan(\pi/5)$, the numerator has argument $\pi/5$ and the denominator is its conjugate.

Their quotient is $\operatorname{cis}(2\pi/5)$. Raising to the tenth power gives $\operatorname{cis}(4\pi)=1$.

The equations $x+1/x=2\cos\alpha$ and $y+1/y=2\cos\beta$ give $x=\operatorname{cis}\alpha$ and $y=\operatorname{cis}\beta$ (choosing the corresponding roots; the reciprocal choices give the same identities).

(i) $x/y+y/x=\operatorname{cis}(\alpha-\beta)+\operatorname{cis}(\beta-\alpha)=2\cos(\alpha-\beta)$.

(ii) $xy-1/(xy)=\operatorname{cis}(\alpha+\beta)-\operatorname{cis}[-(\alpha+\beta)]=2i\sin(\alpha+\beta)$.

Applying the same pair of identities to $x^m/y^n$ and $x^my^n$ gives (iii) and (iv).

z^3 = 27 = 3^3 so z = 3 e^{2πik/3}, k = 0,1,2. Thus z_0 = 3, z_1 = 3 e^{2πi/3} = 3ω, z_2 = 3 e^{4πi/3} = 3ω^2.

z^3 = 8 = 2^3 so z = 2 e^{2πik/3}, k=0,1,2. Hence roots are 2, 2ω, 2ω^2 (with ω^3=1).

The equation z^9 = −1 has solutions z = e^{i(2m+1)π/9}, m=0,1,...,8. Each such z equals cos((2m+1)π/9)+ i sin((2m+1)π/9).

(i) 1+ω = −ω^2 and 1+ω^2 = −ω. So (1+ω)^{128}+(1+ω^2)^{128}=(-ω^2)^{128}+(-ω)^{128}=(-1)^{128}(ω^{256}+ω^{128}). Reduce exponents mod 3: 256≡1,128≡2, so ω^{256}+ω^{128}=ω+ω^2=−1. Hence sum = −1. (ii) Note 2^k mod 3 alternates: 1,2,1,2,... So among k=0..11 there are six exponents ≡1 and six ≡2. Thus sum =6ω+6ω^2=6(ω+ω^2)=6(−1)=−6.

$\dfrac{3}{-1+i}=\dfrac{3(-1-i)}{2}=-\dfrac32-\dfrac32i$. This lies in the third quadrant on the line $y=x$.

Its principal argument is $-3\pi/4$.

$\sin40^\circ+i\cos40^\circ=\cos50^\circ+i\sin50^\circ=\operatorname{cis}50^\circ$.

Its fifth power is $\operatorname{cis}250^\circ=\operatorname{cis}(-110^\circ)$.

Taking modulus squared, $x^2+y^2=|(1+i)(1+2i)\cdots(1+ni)|^2$.

This equals $|1+i|^2|1+2i|^2\cdots|1+ni|^2=2\cdot5\cdot10\cdots(1+n^2)$.

Since $1+\omega=-\omega^2$, $(1+\omega)^7=(-\omega^2)^7=-\omega^{14}=-\omega^2$.

Using $1+\omega+\omega^2=0$, $-\omega^2=1+\omega$. Hence $A=1$ and $B=1$.

$\arg(1+i\sqrt3)=\pi/3$, so the numerator has argument $2\pi/3$.

The denominator has argument $\arg i+\arg(1-i\sqrt3)=\pi/2-\pi/3=\pi/6$.

Therefore the quotient has principal argument $2\pi/3-\pi/6=\pi/2$.

The roots are the non-real cube roots of unity $\omega$ and $\omega^2$. Since $2020\equiv1\pmod3$,

$\alpha^{2020}+\beta^{2020}=\omega+\omega^2=-1$.

Use $-\omega^2-1=\omega$ and $\omega^7=\omega$. The determinant becomes $\begin{vmatrix}1&1&1\\1&\omega&\omega^2\\1&\omega^2&\omega\end{vmatrix}$.

Expanding gives $3(\omega^2-\omega)$. Since $\omega=-1/2+i\sqrt3/2$, $\omega^2-\omega=-i\sqrt3$.

Hence the determinant is $-3\sqrt3i$, so $k=-\sqrt3i$.

$1+\sqrt3i=2\operatorname{cis}(\pi/3)$ and $1-\sqrt3i=2\operatorname{cis}(-\pi/3)$.

Their quotient is $\operatorname{cis}(2\pi/3)$. Its tenth power is $\operatorname{cis}(20\pi/3)=\operatorname{cis}(2\pi/3)$.