241. If $a_i, b_i\in R, i = 1, 2, \ldots, n$, show that $\displaystyle\left(\sum_{i = 1}^na_i\right)^2 + \left(\sum_{i = 1}^nb_i\right)^2 \leq \left(\sum_{n = 1}^n\sqrt{a_i^2 + b_i^2}\right)^2$.
Solution: Consider two complex numbers $z_1 = a_1 + ib_1$ and $z_2 = a_2 + ib_2.$ Now we have to prove $|z_1 + z_2| \le |z_1| + |z_2|$ which can be further extended to prove the result.
$\Rightarrow \sqrt{(a_1 + a_2)^2 + (b_2 + b_2)^2} \le \sqrt{a_1^2 + b_1^2} + \sqrt{a_2^2 + b_2^2}$.
Squaring both sides and simplifying
$\Rightarrow a_1a_2 + b_1b_2 \le \sqrt{(a_1^2 + b_1^2)(a_2^2 + b_2^2)} \Rightarrow (a_1a_2 + b_1b_2)^2 - (a_1^2 + b_1^2)(a_2^2 + b_2^2) \le 0\Rightarrow -(a_1b_2 - a_2b_1)^2 \le 0$.
242. Let $\left|\frac{\overline{z_1} - 2\overline{z_2}}{2 - z_1\overline{z_2}}\right| = 1$ and $|z_2|\neq 1$, where $z_1$ and $z_2$ are complex nubers, show that $|z_1| = 2$.
Solution: Given, $\left|\frac{\overline{z_1} - 2\overline{z_2}}{2 - z_1\overline{z_2}}\right| = 1\Rightarrow |\overline{z_1} - 2\overline{z_2}|^2 = |2 - z_1\overline{z_2}|^2$ $\Rightarrow (\overline{z_1} - 2\overline{z_2})(z_1 - 2z_2) = (2 - z_1\overline{z_2})(2 - \overline{z_1}z_2) \Rightarrow |z_1|^2 - 2z_1\overline{z_2} - 2\overline{z_1}z_2 + 4|z_2|^2 = 4 - 2z_1\overline{z_2} - 2\overline{z_1}z_2+ |z_1|^2|z_2|^2$ $\Rightarrow |z_1|^2|z_2|^2 - 4|z_2|^2 - |z_1|^2 - 4 = 0 \Rightarrow |z_2| = 2\;\because |z_1|\neq 1$.
243. If $z_1$ and $z_2$ are complex numbers and $u = \sqrt{z_1z_2}$, prove that $|z_1| + |z_2| = \left|\frac{z_1 + z_2}{2} + u\right| + \left|\frac{z_1 + z_2}{2} - u\right|$.
Solution:: $\left|\frac{z_1 + z_2}{2} + \sqrt{z_1z_2}\right| + \left|\frac{z_1 + z_2}{2} - \sqrt{z_1z_2}\right|$ $= \frac{1}{2}\left|(\sqrt{z_1} + \sqrt{z_2})^2\right| + \frac{1}{2}\left|(\sqrt{z_1} - \sqrt{z_2})^2\right| = |z_1| + |z_2|$.
244. If $z_1$ and $z_2$ are roots of the equation $\alpha z^2 + 2\beta z + \gamma = 0$, then prove that $|\alpha|(|z_1| + |z_2|) = |\beta + \sqrt{\alpha\gamma}| + |\beta - \sqrt{\alpha\gamma}|$.
Solution: We have proven that $|a + \sqrt{a^2 - b^2}| + |a - \sqrt{a^2 - b^2}| = |a + b| + |a - b|$.
Substituting $a = \beta$ and $b = \sqrt{\alpha\gamma}$ we have $|\beta + \sqrt{\alpha\gamma}| + |\beta - \sqrt{\alpha\gamma}| = |\alpha|\left(|\frac{\beta}{\alpha} + \sqrt{\frac{\gamma}{\alpha}}| +|\frac{\beta}{\alpha} - \sqrt{\frac{\gamma}{\alpha}}|\right)$ $= |\alpha|\left(|-z_1 - z_2 + \sqrt{z_1z_2}| + |-z_1 - z_2 - \sqrt{z_1z_2}|\right) = |\alpha|(|z_1| + |z_2|)$.
245. If $a, b, c$ are complex numbers such that $a + b + c = 0$ and $|a| = |b| = |c| = 1$, find the value of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$.
Solution: We have $|a| = 1 \Rightarrow |a|^2 = 1 \Rightarrow a\overline{a} = 1 \Rightarrow \overline{a} = \frac{1}{a}$. Thus, $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \overline{a} + \overline{b} +\overline{c} = 0 [\because a + b + c = 0]$.
246. If $|z + 4|\leq 3$, find the least and greatest value of $|z + 1|$.
Solution: $|z + 4|\leq 3 \Rightarrow -3 \leq z + 4\leq 6 \Rightarrow 0\leq z + 1\leq 6$.
247. Show that for any two non-zero complex numbers $z_1$ and $z_2, (|z_1| + |z_2|)\left|\frac{z_1}{|z_1|} + \frac{z_2}{|z_2|}\right| \leq 2|z_1 + z_2|$.
Solution: We have to prove that $(|z_1| + |z_2|)\left|\frac{z_1}{|z_1|} + \frac{z_2}{|z_2|}\right| \leq 2|z_1 + z_2|$.
Let $z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$ and $z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$.
Then $(|z_1| + |z_2|)\left|\frac{z_1}{|z_1|} + \frac{z_2}{|z_2|}\right| = (r_1 + r_2)\left|(\cos\theta_1 + \cos\theta_2) + i(\sin\theta_1 + \sin\theta_2)\right| = (r_1 + r_2)\sqrt{2 + 2\cos(\theta_1 - \theta_2)}$
Also, $4|z_1 + z_2|^2 = 4[(r_1\cos\theta_1 + r_2\cos\theta_2)^2 + (r_1\sin\theta_1 + r_2\sin\theta_2)^2] = 4[r_1^2 + r_2^2 + r_1r_2\cos(\theta_1 - \theta_2)]$
and squaring L.H.S. we have $2(r_1 + r_2)^2[1 + \cos(\theta_1 - \theta_2)]^2$. Clearly, L.H.S. $\leq$ R.H.S.
248. Show that the necessary and sufficient condition for both the roots of the equation $z^2 + az + b = 0$ to be unimodular are $|a|\leq 2, |b| = 1$ and $\arg(b) = 2\arg(a)$.
Solution: Given equation is $z^2 + az + b = 0$. Let $p, q$ are two of its roots. Then we have $p + q = -a$ and $pq = b$.
Taking modulus of both we have $|p + q| = |a|$ and $|pq| = b$. Now it is required that $|p| = |q| = 1.$
Therefore. we have $|p + q| \le |p| + |q| = 2 \therefore |a| \le 2.$ Similarly, $|b| = |pq| = |p||q| = 1$. Since $p, q$ have unit modulii, we can have them as $p = cos\theta_1 + isin\theta_1$ and $q = cos\theta_2 + isin\theta_2$.
$\arg(b) = \arg(pq) = \arg(\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)) = \theta_1 + \theta_2$
$\arg(a) = \arg(p + q) = \arg[(\cos\theta_1 + \cos\theta_2) + i(\sin\theta_1 + \sin\theta_2)] = \arg\left[\left(\cos^2\frac{\theta_1}{2} + i^2\sin\frac{\theta_1}{2} + 2i\sin\frac{\theta_1}{2}\cos\frac{\theta_1}{2}\right) + \left(\cos^2\frac{\theta_2}{2} + i^2\sin\frac{\theta_2}{2} +2i\sin\frac{\theta_2}{2}\cos\frac{\theta_2}{2}\right)\right]$
$=\arg\left[\cos\frac{\theta_1 + \theta_2}{2} + i\sin\frac{\theta_1 + \theta_2}{2}\right] = \frac{\theta_1 + \theta_2}{2}$ and hence $\arg(b) = 2\arg(a)$.
249. If $z$ is a complex number, show that $|z|\leq |\Re(z)| + |\Im(z)|\leq \sqrt{2}|z|$.
Solution: Let $z = x + iy$. First we consider first two inequalities $|z| \le |\Re(z)| + |\Im(z)| \Rightarrow \sqrt{x^2 + y^2} \le x + y$. Squaring, we have $x^2 + y^2 \le x^2 + y^2 + 2xy \Rightarrow 2xy \ge 0$,
which is true. Now we consider last two inequalities, $|\Re(z)| + |\Im(z)| \le \sqrt{2}|z|\Rightarrow x + y \le \sqrt{2(x^2 + y^2)}$. Squaring, we have $x^2 + y^2 + 2xy \le 2(x^2 + y^2) \Rightarrow (x - y)^2 \ge 0$, which is also true.
250. If $\left|z - \frac{4}{z}\right| = 2$, show that the greatest value of $|z|$ is $\sqrt{5} + 1$.
Solution: $\left|z - \frac{4}{z}\right| = 2 \Rightarrow |z| - \frac{4}{|z|} \geq 2 \Rightarrow |z|^2 - 2|z| - 4\geq 0$. The greatest root of this equation is $\sqrt{5} + 1$.