Question

Number of solutions of the equation z¹⁰ + 31z⁵-32 = 0 for which Re(z)>0

[JEE ADVANCED-2012]​

Answer 1

Given,

$\small\longrightarrow z^{10}+31z^5-32=0$

Let $\smallk=z^5.$ Then,

$\small\longrightarrow k^2+31k-32=0$

$\small\longrightarrow (k-1)(k+32)=0$

$\small\Longrightarrow k\in\{-32,\ 1\}$

$\small\longrightarrow z^5\in\{-32,\ 1\}$

Consider,

$\small\longrightarrow z^5=-32$

We see that $\small1=1^r=(e^{i2\pi})^r=e^{i2\pi r}.$ (Recall Euler's form, 1 can be a complex number with argument 2π) So the RHS of the equation can be taken as,

$\small\longrightarrow z^5=-32e^{i2\pi r}$

$\small\text{ \longrightarrow z=\left(-32e^{i2\pi r}\right)^{\frac{1}{5}} }$

$\small\text{ \longrightarrow z=-2e^{i\frac{2\pi r}{5}} }$

This represents the 5 complex roots of the equation $\smallz^5=-32$ where r is an integer such that 0 ≤ r ≤ 4, or,

$\small\longrightarrow r\in\{0,\ 1,\ 2,\ 3,\ 4\}$

$\small\text{ \longrightarrow\dfrac{2\pi r}{5}\in\left\{0,\ \dfrac{2\pi}{5},\ \dfrac{4\pi}{5},\ \dfrac{6\pi}{5},\ \dfrac{8\pi}{5}\right\}\quad\dots(1) }$

[Note:- The n roots of the equation $\smallx^n=a$ are in the form $\small\text{ x=a^{\frac{1}{n}}e^{i\frac{2\pi r}{n}} }$ where r is an integer variable such that p ≤ r ≤ q where q - p + 1 = n. Usually r is taken 0 ≤ r ≤ n - 1.]

Taking $\smallz$ in polar form,

$\small\longrightarrow z=-2\cos\left(\dfrac{2\pi r}{5}\right)-2i\sin\left(\dfrac{2\pi r}{5}\right)$

Now,

$\small\longrightarrow Re(z)>0$

$\small\longrightarrow-2\cos\left(\dfrac{2\pi r}{5}\right)>0$

$\small\longrightarrow\cos\left(\dfrac{2\pi r}{5}\right)<0$

$\small\Longrightarrow\dfrac{2\pi r}{5}\in\left(2n\pi+\dfrac{\pi}{2},\ 2n\pi+\dfrac{3\pi}{2}\right),\quad n\in\mathbb{Z}\quad\dots(2)$

Taking (1) ∧ (2),

$\small\text{ \longrightarrow\dfrac{2\pi r}{5}\in\left(\dfrac{\pi}{2},\ \dfrac{3\pi}{2}\right)\cap\left\{0,\ \dfrac{2\pi}{5},\ \dfrac{4\pi}{5},\ \dfrac{6\pi}{5},\ \dfrac{8\pi}{5}\right\}\quad (n=0) }$

$\small\text{ \longrightarrow\dfrac{2\pi r}{5}\in\left\{\dfrac{4\pi}{5},\ \dfrac{6\pi}{5}\right\} }$

$\small\text{ \Longrightarrow z\in\left\{-2e^{i\frac{4\pi}{5}},\ -2e^{i\frac{6\pi}{5}}\right\}\quad\dots(i) }$

Consider,

$\small\longrightarrow z^5=1=e^{i2\pi r}$

$\small\text{ \longrightarrow z=e^{i\frac{2\pi r}{5}} }$

$\small\longrightarrow z=\cos\left(\dfrac{2\pi r}{5}\right)+i\sin\left(\dfrac{2\pi r}{5}\right)$

Now,

$\small\longrightarrow Re(z)>0$

$\small\longrightarrow\cos\left(\dfrac{2\pi r}{5}\right)>0$

$\small\Longrightarrow\dfrac{2\pi r}{5}\in\left(2n\pi-\dfrac{\pi}{2},\ 2n\pi+\dfrac{\pi}{2}\right),\quad n\in\mathbb{Z}\quad\dots(3)$

Taking (1) ∧ (3),

$\small\text{ \longrightarrow\dfrac{2\pi r}{5}\in\left[\left(-\dfrac{\pi}{2},\ \dfrac{\pi}{2}\right)\cup\left(\dfrac{3\pi}{2},\ \dfrac{5\pi}{2}\right)\right]\cap\left\{0,\ \dfrac{2\pi}{5},\ \dfrac{4\pi}{5},\ \dfrac{6\pi}{5},\ \dfrac{8\pi}{5}\right\}\quad (n=0,\ n=1) }$

$\small\text{ \longrightarrow\dfrac{2\pi r}{5}\in\left\{0,\ \dfrac{2\pi}{5},\ \dfrac{8\pi}{5}\right\} }$

$\small\text{ \Longrightarrow z\in\left\{1,\ e^{i\frac{2\pi}{5}},\ e^{i\frac{8\pi}{5}}\right\}\quad\dots(ii) }$

Taking (i) ∨ (ii) we get,

$\small\text{ \longrightarrow z\in\left\{-2e^{i\frac{4\pi}{5}},\ -2e^{i\frac{6\pi}{5}},\ 1,\ e^{i\frac{2\pi}{5}},\ e^{i\frac{8\pi}{5}}\right\} }$

Hence there are 5 solutions.