Question

Let $z$ be a complex number satisfying $\left(z+\frac{1}{z}\right)$

$\left(z+\frac{1}{z}+1\right)=1$. Then the value of

$\left(3z^{100}+\frac{2}{z^{100}}+1\right)\left(z^{100}+\frac{2}{z^{100}}+3\right)$ equals to

• A. 25
• B. 30
• C. 36
• D. 50

Answer 1

Answer: (C)

36

Answer 2

We first note that since z lies on the unit circle (as will be clear later), we can write

$z = e^{i\theta}$ so that $z + \frac{1}{z} = e^{i\theta} + e^{{-i\theta}} = 2\cos\theta$.

The given condition is $(z + \frac{1}{z}) \cdot (z + \frac{1}{z} + 1) = 1$.

Substitute $t = z + \frac{1}{z} = 2\cos\theta$. Then the condition becomes

$t(t + 1) = 1 \implies t^2 + t - 1 = 0$.

One could solve for t, but it is more convenient to return to the trigonometric form: $2\cos\theta (2\cos\theta + 1) = 1 \implies 4\cos^2\theta + 2\cos\theta - 1 = 0$.

Solving this quadratic in $\cos\theta$, we get $\cos\theta = \frac{{-1 \pm \sqrt{5}}}{4}$.

Notice that $\cos\theta = \frac{{\sqrt{5} - 1}}{4} \approx 0.309$ corresponds to $\theta = 72^\circ$ (or equivalently $\frac{2\pi}{5}$ radians)

and $\cos\theta = \frac{{-\sqrt{5} - 1}}{4}$ gives another valid angle ($144^\circ$). In either case, since $|z| = 1$, when we raise z to a high power the exponent will effectively be reduced modulo $360^\circ$.

Consider $z^{100} = e^{i100\theta}$. For $\theta = 72^\circ$, we have:

$100\theta = 100 \times 72^\circ = 7200^\circ$. Since $7200^\circ/360^\circ = 20$ exactly, it follows that

$z^{100} = e^{i7200^\circ} = 1$.

Similarly, for the other value of $\theta$, $z^{100}$ will also be 1 because $100\theta$ will be an integral multiple of $360^\circ$.

Now the given expression is $(3z^{100} + \frac{2}{z^{100}} + 1) \cdot (z^{100} + \frac{2}{z^{100}} + 3)$.

Since $z^{100} = 1$ and $\frac{1}{z^{100}} = 1$, the expression reduces to

$(3\cdot1 + 2\cdot1 + 1) \cdot (1 + 2\cdot1 + 3) = (3 + 2 + 1)(1 + 2 + 3) = 6\cdot6 = 36$.

Thus, the value is 36.