Question

Let z is a complex number satisfying the equation ${{z}^{6}}+{{z}^{3}}+1=0$. If this equation has a root $r{{e}^{i\theta }}\,with \,{{90}^{{}^\circ }}<\theta \,<\,{{180}^{{}^\circ }}$ then the value of ‘theta’ is
(a) ${{100}^{{}^\circ }}$
(b) ${{110}^{{}^\circ }}$
(c) ${{160}^{{}^\circ }}$
(d) ${{170}^{{}^\circ }}$

Answer 1

Hint: Use the Euler’s formula of sin; cos of a particular angle. Here the left-hand side can also be written as cisx its own wish as both are the same.
${{e}^{ix}}=\cos x+i\sin x$

Complete step-by-step solution -
Given expression in the question of the variable Z is:
$Z=r{{e}^{i\theta }}$
For our easy representation we take $\theta $ as q in our solution.
$Z=r{{e}^{iq}}$
By Euler’s formula of sin, cos of an angle q is:
${{e}^{iq}}=\cos q+i\sin q$
By substituting this into our term Z, we term Z into:
$Z=r\left( \cos q+i\sin q \right)$
Given equation in the question for which we need to find the solution:
${{z}^{6}}+{{z}^{3}} + 1=0$
So, for our convenience, we can assume a variable k.
The k satisfies the condition, in terms of the number Z:
$k={{z}^{3}}$
By substituting this k value into the equation, we get:
${{k}^{2}}+k+1=0$
By basic knowledge of algebra, we can say that the:
If equation $a{{x}^{2}}+bx+c=0$, is true then root of equation can be:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
By applying this here, we get value of k to be:
$k=\dfrac{-1\pm \sqrt{1-4}}{2\left( 1 \right)}$
By simplifying this, we get value of k to be
$k=\dfrac{-1\pm i\sqrt{3}}{2}$
So, finally we get the values of k to be as:
$k=\dfrac{-1+i\sqrt{3}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,k=\dfrac{-1-i\sqrt{3}}{2}$
Case 1: Take first value of k and find the angle asked
$k=\dfrac{-1+i\sqrt{3}}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,$
By using Euler’s formula, we can convert k into:
$Z=\cos \theta +i\sin \theta ={{e}^{i\theta }}$
Here angle is $\dfrac{2\pi }{3}$, so k can be written as:
$k={{e}^{i\left( \dfrac{2\pi }{3} \right)}}$
By substituting k value back, we get Z as:
${{Z}^{3}}={{e}^{i\left( \dfrac{2\pi }{3} \right)}}\,\,\,\,\,\,\,\,given\,Z={{e}^{i\theta }}$
${{e}^{3i\theta }}={{e}^{i\left( \dfrac{2\pi }{3} \right)}}\Rightarrow 3\theta =2n\pi +\dfrac{2\pi }{3}..........(i)$
Case 2: $k=\dfrac{-1-i\sqrt{3}}{2}$
By using this formula, here angle B $\dfrac{4\pi }{3}$, we get:
$k={{e}^{i\left( \dfrac{4\pi }{3} \right)}}$
Similarly, as above, ${{e}^{3i\theta }}={{e}^{i\left( \dfrac{4\pi }{3} \right)}}\Rightarrow 3\theta =2n\pi +\dfrac{4\pi }{3}..........(ii)$
By equation (i) and equation (ii) the possible values of angles are obtained by substituting $n=1,2,.........$ by this, we get:
$\theta =40,80,160,200,280,320,.......$
By options we select $\theta =160{}^\circ $
So, option (c) is correct.

Note: Be careful while getting ${{2}^{3}}$ into the equation, don’t forget that $k={{2}^{3}}$$\Rightarrow 3\theta $ should come into play not $\theta $. So, the angle you get by k will be divided by 3. From equation (i) and (ii) common angles will be a suitable angle out of them which angle matches with options will be our solution.