Mix Examples-Complex numbers Questions in English

(C) Since the coefficients $a, b, c, d$ are real,complex roots must occur in conjugate pairs.
Given roots are $z_1 = 2+i$ and $z_3 = \sqrt{5}-2i$.
Therefore,their conjugates $z_2 = 2-i$ and $z_4 = \sqrt{5}+2i$ must also be roots of the equation.
The product of all roots is given by $z_1 \times z_2 \times z_3 \times z_4$.
Product $= (2+i)(2-i) \times (\sqrt{5}-2i)(\sqrt{5}+2i)$.
Using the identity $(x+iy)(x-iy) = x^2+y^2$:
Product $= (2^2+1^2) \times ((\sqrt{5})^2+2^2) = (4+1) \times (5+4) = 5 \times 9 = 45$.

(C) Given expression: $\frac{(1+i)^{n}}{(1-i)^{n-2}}$
$= \frac{(1+i)^{n}}{(1-i)^{n} \cdot (1-i)^{-2}}$
$= \left(\frac{1+i}{1-i}\right)^{n} \cdot (1-i)^{2}$
$= \left(\frac{(1+i)(1+i)}{(1-i)(1+i)}\right)^{n} \cdot (1 + i^{2} - 2i)$
$= \left(\frac{1 + 2i + i^{2}}{1 - i^{2}}\right)^{n} \cdot (1 - 1 - 2i)$
$= \left(\frac{1 + 2i - 1}{1 + 1}\right)^{n} \cdot (-2i)$
$= \left(\frac{2i}{2}\right)^{n} \cdot (-2i)$
$= i^{n} \cdot (-2i)$
$= -2i^{n+1}$

(D) Given $z^{1/3} = p + iq$,we have $z = (p + iq)^3$.
Expanding this,$z = p^3 + 3p^2(iq) + 3p(iq)^2 + (iq)^3 = p^3 + 3ip^2q - 3pq^2 - iq^3$.
Grouping real and imaginary parts,$z = (p^3 - 3pq^2) + i(3p^2q - q^3)$.
Since $z = x - iy$,we equate the real and imaginary parts:
$x = p^3 - 3pq^2 = p(p^2 - 3q^2)$
$-y = 3p^2q - q^3 \implies y = q^3 - 3p^2q = -q(3p^2 - q^2)$.
Now,calculate $\frac{x}{p} + \frac{y}{q}$:
$\frac{x}{p} = p^2 - 3q^2$
$\frac{y}{q} = q^2 - 3p^2$
$\frac{x}{p} + \frac{y}{q} = (p^2 - 3q^2) + (q^2 - 3p^2) = -2p^2 - 2q^2 = -2(p^2 + q^2)$.
Finally,$\frac{(\frac{x}{p} + \frac{y}{q})}{(p^2 + q^2)} = \frac{-2(p^2 + q^2)}{(p^2 + q^2)} = -2$.

(A) Given that $\frac{2 z_1}{3 z_2}$ is a purely imaginary number,we have $\frac{z_1}{z_2} = i k$ for some real number $k \neq 0$.
We want to find the value of $\left|\frac{z_1-z_2}{z_1+z_2}\right|$.
Dividing the numerator and denominator by $z_2$,we get:
$\left|\frac{\frac{z_1}{z_2}-1}{\frac{z_1}{z_2}+1}\right| = \left|\frac{i k-1}{i k+1}\right|$.
Since the modulus of a quotient is the quotient of the moduli,we have:
$\frac{|i k-1|}{|i k+1|} = \frac{\sqrt{(-1)^2 + k^2}}{\sqrt{1^2 + k^2}} = \frac{\sqrt{1+k^2}}{\sqrt{1+k^2}} = 1$.

(A) Let $z = \frac{1-i \cos \theta}{1+2 i \cos \theta}$.
Since $z$ is a real number,$z = \bar{z}$.
$\frac{1-i \cos \theta}{1+2 i \cos \theta} = \frac{1+i \cos \theta}{1-2 i \cos \theta}$.
Cross-multiplying,we get:
$(1-i \cos \theta)(1-2 i \cos \theta) = (1+i \cos \theta)(1+2 i \cos \theta)$.
$1 - 2i \cos \theta - i \cos \theta + 2i^2 \cos^2 \theta = 1 + 2i \cos \theta + i \cos \theta + 2i^2 \cos^2 \theta$.
Since $i^2 = -1$,this simplifies to:
$1 - 3i \cos \theta - 2 \cos^2 \theta = 1 + 3i \cos \theta - 2 \cos^2 \theta$.
Subtracting common terms from both sides:
$-3i \cos \theta = 3i \cos \theta$.
$6i \cos \theta = 0$.
Since $i \neq 0$,we must have $\cos \theta = 0$.
Therefore,$\theta = (2n+1) \frac{\pi}{2}$ for $n \in I$.

(A) Given $|z_{1}|=|z_{2}|=|z_{3}|=1$.
Since $|z|=1$ $\Rightarrow z\bar{z}=1$ $\Rightarrow \bar{z}=\frac{1}{z}$.
Thus,$\frac{1}{z_{1}}=\bar{z}_{1}$,$\frac{1}{z_{2}}=\bar{z}_{2}$,and $\frac{1}{z_{3}}=\bar{z}_{3}$.
We are given $|\frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}}|=1$.
Substituting the conjugates,we get $|\bar{z}_{1}+\bar{z}_{2}+\bar{z}_{3}|=1$.
Using the property $|\bar{z}|=|z|$,we have $|\overline{z_{1}+z_{2}+z_{3}}|=1$.
Therefore,$|z_{1}+z_{2}+z_{3}|=1$.

(B) We have,$|z| = \left|z-\frac{3}{z}+\frac{3}{z}\right|$
Using the triangle inequality,$|z| \leq \left|z-\frac{3}{z}\right| + \left|\frac{3}{z}\right|$
Given $\left|z-\frac{3}{z}\right| = 2$,we get $|z| \leq 2 + \frac{3}{|z|}$
Multiplying by $|z|$ (since $|z| > 0$),we have $|z|^2 \leq 2|z| + 3$
$|z|^2 - 2|z| - 3 \leq 0$
$(|z|-3)(|z|+1) \leq 0$
Since $|z| \geq 0$,we must have $|z| \leq 3$
Thus,the maximum value of $|z|$ is $3$.

(C) Given the condition $\left|z+\frac{2}{z}\right|=2$.
Using the triangle inequality,$|z| = \left|z+\frac{2}{z}-\frac{2}{z}\right| \leq \left|z+\frac{2}{z}\right| + \left|-\frac{2}{z}\right|$.
Substituting the given value,$|z| \leq 2 + \frac{2}{|z|}$.
Multiplying by $|z|$ (since $|z| > 0$),we get $|z|^2 \leq 2|z| + 2$.
Rearranging the terms,$|z|^2 - 2|z| \leq 2$.
Completing the square,$|z|^2 - 2|z| + 1 \leq 2 + 1$,which gives $(|z|-1)^2 \leq 3$.
Taking the square root,$-\sqrt{3} \leq |z|-1 \leq \sqrt{3}$.
Adding $1$ to all sides,$1-\sqrt{3} \leq |z| \leq 1+\sqrt{3}$.
Since $|z| \geq 0$,the maximum value of $|z|$ is $1+\sqrt{3}$.

(B) Given that $P, Q$ and $R$ are angles of an isosceles triangle and $\angle P = \frac{\pi}{2}$.
Since $P + Q + R = \pi$,we have $Q + R = \pi - \frac{\pi}{2} = \frac{\pi}{2}$.
Since the triangle is isosceles and $\angle P = \frac{\pi}{2}$,we have $Q = R = \frac{\pi}{4}$.
Using the identity $\cos \theta + i \sin \theta = e^{i \theta}$,the expression becomes:
$E = (e^{-i P/3})^3 + (e^{i Q})(e^{-i R}) + (e^{-i P})(e^{-i Q})(e^{-i R})$
$E = e^{-i P} + e^{i(Q - R)} + e^{-i(P + Q + R)}$
Substituting $P = \frac{\pi}{2}, Q = \frac{\pi}{4}, R = \frac{\pi}{4}$:
$E = e^{-i \pi/2} + e^{i(0)} + e^{-i(\pi/2 + \pi/4 + \pi/4)}$
$E = -i + 1 + e^{-i \pi}$
$E = -i + 1 - 1 = -i$.

(D) Given the quadratic equation $z^2 + 4z - (1 + 12i) = 0$.
Using the quadratic formula $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,where $a = 1, b = 4, c = -(1 + 12i)$:
$z = \frac{-4 \pm \sqrt{16 - 4(1)(-(1 + 12i))}}{2} = \frac{-4 \pm \sqrt{16 + 4 + 48i}}{2} = \frac{-4 \pm \sqrt{20 + 48i}}{2} = -2 \pm \sqrt{5 + 12i}$.
Let $\sqrt{5 + 12i} = x + iy$. Squaring both sides: $x^2 - y^2 + 2ixy = 5 + 12i$.
Equating real and imaginary parts: $x^2 - y^2 = 5$ and $2xy = 12 \Rightarrow xy = 6$.
Since $(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2 = 5^2 + 12^2 = 25 + 144 = 169$,we have $x^2 + y^2 = 13$.
Solving $x^2 - y^2 = 5$ and $x^2 + y^2 = 13$,we get $2x^2 = 18 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3$.
If $x = 3, y = 2$; if $x = -3, y = -2$. Thus $\sqrt{5 + 12i} = \pm(3 + 2i)$.
So,$z = -2 \pm (3 + 2i)$.
$z_1 = -2 + 3 + 2i = 1 + 2i$ and $z_2 = -2 - 3 - 2i = -5 - 2i$.
$|z_1|^2 = 1^2 + 2^2 = 1 + 4 = 5$.
$|z_2|^2 = (-5)^2 + (-2)^2 = 25 + 4 = 29$.
$|z_1|^2 + |z_2|^2 = 5 + 29 = 34$.

(A) Let $z = x+iy$. The equation is $z\bar{z} + z(2+i) + k(2+3i) = 0$.
Substituting $z = x+iy$ and $\bar{z} = x-iy$,we get $(x^2 + y^2) + (x+iy)(2+i) + 2k + 3ki = 0$.
Expanding the terms: $x^2 + y^2 + 2x + ix + 2iy - y + 2k + 3ki = 0$.
Separating real and imaginary parts:
Real part: $x^2 + y^2 + 2x - y + 2k = 0$
Imaginary part: $x + 2y + 3k = 0 \Rightarrow x = -2y - 3k$.
Substitute $x$ into the real part equation:
$(-2y-3k)^2 + y^2 + 2(-2y-3k) - y + 2k = 0$
$4y^2 + 12yk + 9k^2 + y^2 - 4y - 6k - y + 2k = 0$
$5y^2 + y(12k - 5) + 9k^2 - 4k = 0$.
For $y$ to be real,the discriminant $D \ge 0$:
$D = (12k - 5)^2 - 4(5)(9k^2 - 4k) \ge 0$
$144k^2 - 120k + 25 - 180k^2 + 80k \ge 0$
$-36k^2 - 40k + 25 \ge 0 \Rightarrow 36k^2 + 40k - 25 \le 0$.
The roots of $36k^2 + 40k - 25 = 0$ are $k = \frac{-40 \pm \sqrt{1600 - 4(36)(-25)}}{72} = \frac{-40 \pm \sqrt{5200}}{72}$.
Thus,$\alpha + \beta = -\frac{40}{36} = -\frac{10}{9}$.
Therefore,$9(\alpha + \beta) = 9(-\frac{10}{9}) = -10$.

(D) The roots of the quadratic equation $z^2 + 4z + 16 = 0$ are found using the quadratic formula $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
Substituting the values $a = 1, b = 4, c = 16$,we get $z = \frac{-4 \pm \sqrt{16 - 64}}{2} = \frac{-4 \pm \sqrt{-48}}{2} = -2 \pm i 2\sqrt{3}$.
Let the two roots be $z_1 = -2 + 2\sqrt{3}i$ and $z_2 = -2 - 2\sqrt{3}i$.
We need to calculate the sum $S = |z_1 + \sqrt{3}i|^2 + |z_2 + \sqrt{3}i|^2$.
First,calculate $|z_1 + \sqrt{3}i|^2 = |-2 + 2\sqrt{3}i + \sqrt{3}i|^2 = |-2 + 3\sqrt{3}i|^2 = (-2)^2 + (3\sqrt{3})^2 = 4 + 27 = 31$.
Next,calculate $|z_2 + \sqrt{3}i|^2 = |-2 - 2\sqrt{3}i + \sqrt{3}i|^2 = |-2 - \sqrt{3}i|^2 = (-2)^2 + (-\sqrt{3})^2 = 4 + 3 = 7$.
Finally,the sum is $31 + 7 = 38$.

(A) Given the quadratic equation $z^2 + \sqrt{6}iz - 3 = 0$.
Using the quadratic formula $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get:
$z = \frac{-\sqrt{6}i \pm \sqrt{(\sqrt{6}i)^2 - 4(1)(-3)}}{2} = \frac{-\sqrt{6}i \pm \sqrt{-6 + 12}}{2} = \frac{-\sqrt{6}i \pm \sqrt{6}}{2}$.
Thus,the roots are $z_1 = \frac{\sqrt{6}}{2}(1 - i)$ and $z_2 = \frac{\sqrt{6}}{2}(-1 - i)$.
Now,calculate $z^2$ for each root:
$z_1^2 = \left(\frac{\sqrt{6}}{2}(1 - i)\right)^2 = \frac{6}{4}(1 - 2i + i^2) = \frac{3}{2}(1 - 2i - 1) = -3i$.
$z_2^2 = \left(\frac{\sqrt{6}}{2}(-1 - i)\right)^2 = \frac{6}{4}(1 + 2i + i^2) = \frac{3}{2}(1 + 2i - 1) = 3i$.
Now,calculate $z^8 = (z^2)^4$:
$z_1^8 = (-3i)^4 = (-3)^4 \cdot i^4 = 81 \cdot 1 = 81$.
$z_2^8 = (3i)^4 = 3^4 \cdot i^4 = 81 \cdot 1 = 81$.
Therefore,$\sum_{z \in S} z^8 = z_1^8 + z_2^8 = 81 + 81 = 162$.