Conjugate and Modulus of complex numbers Questions in English

(B) Given,$z = a - \frac{i}{2}$.
We need to calculate $|i + z|^2 - |i - z|^2$.
First,substitute $z$ into the expressions:
$i + z = i + a - \frac{i}{2} = a + \frac{i}{2}$.
$i - z = i - (a - \frac{i}{2}) = -a + \frac{3i}{2}$.
Now,calculate the squared moduli:
$|i + z|^2 = |a + \frac{i}{2}|^2 = a^2 + (\frac{1}{2})^2 = a^2 + \frac{1}{4}$.
$|i - z|^2 = |-a + \frac{3i}{2}|^2 = (-a)^2 + (\frac{3}{2})^2 = a^2 + \frac{9}{4}$.
Finally,subtract the two values:
$|i + z|^2 - |i - z|^2 = (a^2 + \frac{1}{4}) - (a^2 + \frac{9}{4}) = \frac{1}{4} - \frac{9}{4} = -\frac{8}{4} = -2$.

(C) Let $z_1 = 1+4i, z_2 = 3+i, z_3 = 1-i$ and $z_4 = 2-3i$.
The modulus of a complex number $z = a+bi$ is given by $|z| = \sqrt{a^2+b^2}$.
Calculating the moduli:
$m_1 = |1+4i| = \sqrt{1^2+4^2} = \sqrt{1+16} = \sqrt{17}$
$m_2 = |3+i| = \sqrt{3^2+1^2} = \sqrt{9+1} = \sqrt{10}$
$m_3 = |1-i| = \sqrt{1^2+(-1)^2} = \sqrt{1+1} = \sqrt{2}$
$m_4 = |2-3i| = \sqrt{2^2+(-3)^2} = \sqrt{4+9} = \sqrt{13}$
Comparing the values: $\sqrt{2} < \sqrt{10} < \sqrt{13} < \sqrt{17}$,which implies $m_3 < m_2 < m_4 < m_1$.

(C) Given the complex numbers:
$\alpha_1 = |-i| = 1$
$\alpha_2 = |\frac{1}{3}(1+i)| = \frac{1}{3} \sqrt{1^2 + 1^2} = \frac{\sqrt{2}}{3} \approx 0.471$
$\alpha_3 = |-1+i| = \sqrt{(-1)^2 + 1^2} = \sqrt{2} \approx 1.414$
Comparing the values: $\frac{\sqrt{2}}{3} < 1 < \sqrt{2}$.
Therefore,the increasing order is $\alpha_2 < \alpha_1 < \alpha_3$.

(A) Given that $\left|\frac{z_1-3 z_2}{3-z_1 \bar{z}_2}\right|=1$ and $\left|z_1\right| \neq 3$.
Squaring both sides,we get $\left|z_1-3 z_2\right|^2 = \left|3-z_1 \bar{z}_2\right|^2$.
Using the property $|z|^2 = z \bar{z}$,we have $(z_1-3 z_2)(\bar{z}_1-3 \bar{z}_2) = (3-z_1 \bar{z}_2)(3-\bar{z}_1 z_2)$.
Expanding both sides: $|z_1|^2 - 3z_1 \bar{z}_2 - 3z_2 \bar{z}_1 + 9|z_2|^2 = 9 - 3\bar{z}_1 z_2 - 3z_1 \bar{z}_2 + |z_1|^2 |z_2|^2$.
Canceling common terms $-3z_1 \bar{z}_2$ and $-3z_2 \bar{z}_1$ from both sides: $|z_1|^2 + 9|z_2|^2 = 9 + |z_1|^2 |z_2|^2$.
Rearranging the terms: $|z_1|^2 - 9 - |z_1|^2 |z_2|^2 + 9|z_2|^2 = 0$.
Factorizing: $(|z_1|^2 - 9) - |z_2|^2 (|z_1|^2 - 9) = 0$.
$(|z_1|^2 - 9)(1 - |z_2|^2) = 0$.
Since $|z_1| \neq 3$,we must have $|z_2|^2 = 1$,which implies $|z_2| = 1$.

(C) The equation of a circle with center at the origin and radius $r$ is given by $|z| = r$.
Given $z = (1+i)(1+2i)(1+3i) \ldots (1+10i)$.
Taking the modulus on both sides:
$|z| = |1+i| \cdot |1+2i| \cdot |1+3i| \ldots |1+10i| = r$.
Using the property $|a+bi| = \sqrt{a^2+b^2}$,we get:
$|z| = \sqrt{1^2+1^2} \cdot \sqrt{1^2+2^2} \cdot \sqrt{1^2+3^2} \ldots \sqrt{1^2+10^2} = r$.
$|z| = \sqrt{2} \cdot \sqrt{5} \cdot \sqrt{10} \ldots \sqrt{101} = r$.
Squaring both sides,we obtain:
$r^2 = 2 \times 5 \times 10 \times \ldots \times 101$.

(C) First,simplify the expression: $\frac{1-i}{3+i} + \frac{4i}{5} = \frac{5(1-i) + 4i(3+i)}{5(3+i)}$
$= \frac{5 - 5i + 12i + 4i^2}{5(3+i)}$
Since $i^2 = -1$,we have $\frac{5 + 7i - 4}{15 + 5i} = \frac{1 + 7i}{15 + 5i}$
Multiply the numerator and denominator by the conjugate $(15 - 5i)$:
$= \frac{(1 + 7i)(15 - 5i)}{(15 + 5i)(15 - 5i)} = \frac{15 - 5i + 105i - 35i^2}{225 + 25} = \frac{15 + 100i + 35}{250} = \frac{50 + 100i}{250} = \frac{1 + 2i}{5}$
Now,find the modulus: $|\frac{1}{5} + \frac{2}{5}i| = \sqrt{(\frac{1}{5})^2 + (\frac{2}{5})^2} = \sqrt{\frac{1}{25} + \frac{4}{25}} = \sqrt{\frac{5}{25}} = \frac{\sqrt{5}}{5}$ unit.

(D) Given $z = \frac{4}{1-i}$.
To find the conjugate $\bar{z}$,we take the conjugate of the expression:
$\bar{z} = \overline{\left(\frac{4}{1-i}\right)} = \frac{\bar{4}}{\overline{1-i}}$.
Since the conjugate of a real number $4$ is $4$ and the conjugate of $(1-i)$ is $(1+i)$,
$\bar{z} = \frac{4}{1+i}$.

(B) We are given $z = \prod_{r=1}^n (1+ri)$.
Taking the modulus squared on both sides,we get $|z|^2 = \prod_{r=1}^n |1+ri|^2$.
Since $|1+ri|^2 = 1^2 + r^2 = 1+r^2$,we have $|z|^2 = \prod_{r=1}^n (1+r^2) = 44200$.
For $n=1$: $1+1^2 = 2$.
For $n=2$: $2 \times (1+2^2) = 2 \times 5 = 10$.
For $n=3$: $10 \times (1+3^2) = 10 \times 10 = 100$.
For $n=4$: $100 \times (1+4^2) = 100 \times 17 = 1700$.
For $n=5$: $1700 \times (1+5^2) = 1700 \times 26 = 44200$.
Thus,$n=5$.