(a) Solve the equation $z^3+8i=0$, where $z\in\mathbb{C}$. **OR** (b) Solve : $\left(1+x+xy^2\right)\dfrac{dy}{dx}+\left(y+y^3\right)=0$.

(a) Solve the equation z³+8i=0, where z∈mathbbC. **OR** (b) Solve : (1+x+xy²)(dy)/(dx)+(y+y³)=0.

Question 109 of 122

z^3+8i=0

(b) Solve :

\left(1+x+xy^2\right)\dfrac{dy}{dx}+\left(y+y^3\right)=0

Puducherry TnboardTamil Nadu HSC (DGE) Board 2023Subjective· 5mImportance★★★★★

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Concept understanding — Roots of Complex Numbers / nth Roots of Unity

$n$ th roots of a complex number. If $\omega=\rho(\cos\phi+i\sin\phi)$ satisfies $\omega^n=z$ where $z=r(\cos\theta+i\sin\theta)$ , write $z$ with its general argument $\theta+2k\pi$ (since a single value of $\theta$ would miss roots) and apply de Moivre's theorem to $\omega^n$ : comparing modulus and argument gives $\rho^n=r$ and $n\phi=\theta+2k\pi$ , so

\rho=r^{1/n},\qquad \phi=\frac{\theta+2k\pi}{n}.

Hence the $n$ th roots of $z=r(\cos\theta+i\sin\theta)$ are

z^{1/n}=r^{1/n}\left(\cos\frac{\theta+2k\pi}{n}+i\sin\frac{\theta+2k\pi}{n}\right),\qquad k=0,1,2,\dots,n-1.

Although $k$ can be any integer, only $k=0,1,\dots,n-1$ give distinct values (larger $k$ repeats the same $n$ roots cyclically). Geometrically, all $n$ roots share the modulus $r^{1/n}$ , so they lie on a circle of radius $r^{1/n}$ centred at the origin, equally spaced at angular intervals of $\dfrac{2\pi}n$ — i.e. at the vertices of a regular $n$ -gon.

$n$ th roots of unity. Setting $z=1=\cos0+i\sin0$ specialises the formula to

z=\cos\frac{2k\pi}n+i\sin\frac{2k\pi}n=e^{2k\pi i/n},\qquad k=0,1,\dots,n-1.

Writing $\omega=e^{2\pi i/n}$ (the primitive $n$ th root), the $n$ roots are exactly $1,\omega,\omega^2,\dots,\omega^{n-1}$ — a geometric progression with common ratio $\omega$ — and they are the vertices of a regular $n$ -gon inscribed in the unit circle.

Standing facts about the $n$ th roots of unity (all provable from the GP sum/product formulas):

Sum $1+\omega+\omega^2+\cdots+\omega^{n-1}=0$ (a finite GP with ratio $\omega\ne1$ , sum $\dfrac{\omega^n-1}{\omega-1}=\dfrac{1-1}{\omega-1}=0$ ).
Product $1\cdot\omega\cdot\omega^2\cdots\omega^{n-1}=(-1)^{n-1}$ .
They all satisfy $|z|=1$ and $z^n=1$ . …

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(a) Solves $z^3+8i=0$ using De Moivre's theorem on the polar form of $-8i$ ; (b) turns the given ODE into a linear equation for $x$ as a function of $y$ and solves via an integrating factor. Both alternatives answered below.

(a) Solve $z^3+8i=0$

1. Rewrite in polar form. $z^3=-8i$ . Now $-8i$ has modulus $8$ and lies on the negative imaginary axis, so its principal argument is $-\dfrac\pi2$ :

$-8i=8\left[\cos\left(-\dfrac\pi2\right)+i\sin\left(-\dfrac\pi2\right)\right]$

2. General polar form. Since adding $2k\pi$ to the argument doesn't change the complex number:

$z^3=8\left[\cos\left(-\dfrac\pi2+2k\pi\right)+i\sin\left(-\dfrac\pi2+2k\pi\right)\right],\quad k=0,1,2$

3. Apply De Moivre's theorem (cube root).

$z=8^{1/3}\left[\cos\left(\dfrac{-\pi/2+2k\pi}{3}\right)+i\sin\left(\dfrac{-\pi/2+2k\pi}{3}\right)\right]=2\,\text{cis}\left(\dfrac{-\pi/2+2k\pi}{3}\right)$

4. Compute each root.

$k=0$ : angle $=-\dfrac\pi6$ . $z=2\left(\cos\left(-\dfrac\pi6\right)+i\sin\left(-\dfrac\pi6\right)\right)=2\left(\dfrac{\sqrt3}{2}-\dfrac i2\right)=\sqrt3-i$ .
$k=1$ : angle $=\dfrac{-\pi/2+2\pi}{3}=\dfrac{3\pi/2}{3}=\dfrac\pi2$ . $z=2\left(\cos\dfrac\pi2+i\sin\dfrac\pi2\right)=2i$ .
$k=2$ : angle $=\dfrac{-\pi/2+4\pi}{3}=\dfrac{7\pi/2}{3}=\dfrac{7\pi}{6}$ . $z=2\left(\cos\dfrac{7\pi}6+i\sin\dfrac{7\pi}6\right)=2\left(-\dfrac{\sqrt3}2-\dfrac i2\right)=-\sqrt3-i$ .

5. Check (e.g. $k=1$ ): $(2i)^3=8i^3=8(-i)=-8i$ , so $z^3+8i=-8i+8i=0$ . ✓

So the three roots are $z=\sqrt3-i,\ 2i,\ -\sqrt3-i$ .

(b) Solve $(1+x+xy^2)\dfrac{dy}{dx}+(y+y^3)=0$

1. Factor. Write $1+x+xy^2=1+x(1+y^2)$ and $y+y^3=y(1+y^2)$ . The equation becomes

$[1+x(1+y^2)]\dfrac{dy}{dx}=-y(1+y^2)$

2. Switch to treating $x$ as a function of $y$ . Taking reciprocals,

$\dfrac{dx}{dy}=\dfrac{1+x(1+y^2)}{-y(1+y^2)}=-\dfrac{1}{y(1+y^2)}-\dfrac{x}{y}$

3. Rearrange into standard linear form. …

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Q.(a) Solve the equation z3+8i=0z^3+8i=0z3+8i=0, where z∈Cz\in\mathbb{C}z∈C. OR

(a) Solve z3+8i=0z^3+8i=0z3+8i=0

(b) Solve (1+x+xy2)dydx+(y+y3)=0(1+x+xy^2)\dfrac{dy}{dx}+(y+y^3)=0(1+x+xy2)dxdy​+(y+y3)=0

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Q.(a) Solve the equation $z^3+8i=0$ , where $z\in\mathbb{C}$ . OR

(a) Solve $z^3+8i=0$

(b) Solve $(1+x+xy^2)\dfrac{dy}{dx}+(y+y^3)=0$

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