The Absolute Value (Modulus) of a Complex Number

In the complex plane, the distance from the origin to the point corresponding to a complex number is called the absolute value or modulus of that complex number. The modulus of the number \( z = a \ \ + \ \ bi \) is denoted by \( |z| \).

In the complex plane, the modulus of \( z = a \ \ + \ \ bi \) is the distance of the point corresponding to this number from the origin.

By the Pythagorean theorem, this distance satisfies:

\[
|z|^{2} = a^{2} \ \ + \ \ b^{2}
\]

Therefore, it is evaluated as:

\[
|z| = \sqrt{ a^{2} \ \ + \ \ b^{2} }
\]

Examples:

\( z = -3 \ \ + \ \ 4i \)

\( |z| = \sqrt{ (-3)^{2} \ \ + \ \ 4^{2} } = 5 \)

\( z = 1 \ \ – \ \ 2\sqrt{2}i \)

\( |z| = \sqrt{ 1^{2} \ \ + \ \ (-2\sqrt{2})^{2} } = 3 \)

\( z = 1 \ \ + \ \ \sqrt{2} \ \ – \ \ i \)

\( |z| = \sqrt{ (1 \ \ + \ \ \sqrt{2})^{2} \ \ + \ \ (-1)^{2} } = \sqrt{ 4 \ \ + \ \ 2\sqrt{2} } \)

\( z = -3i \)

\( |z| = \sqrt{ 0^{2} \ \ + \ \ (-3)^{2} } = 3 \)

Properties of the Modulus:

1) \( |z| = |-z| = |\bar{z}| = |-\bar{z}| \)

2) \( |z_{1} \cdot z_{2}| = |z_{1}| \cdot |z_{2}| \)

3) \( \left| \displaystyle \frac{ z_{1} }{ z_{2} } \right| = \displaystyle \frac{ |z_{1}| }{ |z_{2}| } \) ( where \( z_{2} \neq 0 \) )

4) \( |z^{n}| = |z|^{n} \)

5) \( z \cdot \bar{z} = |z|^{2} \)

6) \( |z_{1} \ \ + \ \ z_{2}| \leq |z_{1}| \ \ + \ \ |z_{2} \quad \) (Triangle Inequality)

7) \( |z_{1} \ \ – \ \ z_{2}| \geq \left| |z_{1}| \ \ – \ \ |z_{2}| \right| \)

Examples:

\( \left| \displaystyle \frac{ 2 \ \ + \ \ \sqrt{5}i }{ 1 \ \ + \ \ 2\sqrt{2}i } \right| \)

\( = \displaystyle\frac{ \sqrt{ 2^{2} \ \ + \ \ (\sqrt{5})^{2} } }{ \sqrt{ 1^{2} \ \ + \ \ (2\sqrt{2})^{2} } } \)

\( =\displaystyle \frac{ 3 }{ 3 } = 1 \)

\( |(\sqrt{3} \ \ – \ \ i)^{7} \cdot (3 \ \ + \ \ 4i)| \)

\( = |(\sqrt{3} \ \ – \ \ i)^{7}| \cdot |3 \ \ + \ \ 4i| \)

\( = \left( \sqrt{ (\sqrt{3})^{2} \ \ + \ \ (-1)^{2} } \right)^{7} \cdot 5 \)

\( = 2^{7} \cdot 5 = 640 \)

QUESTION 18

What is the modulus of the complex number \( z = \displaystyle \frac{ (-8 \ \ + \ \ 6i)^{5} }{ \sqrt[3]{ 7 \ \ + \ \ \sqrt{15}i } } \)?

\[ A) \ 5 \cdot 10^{3} \quad B) \ 5 \cdot 10^{4} \quad C) \ 5 \cdot 10^{5} \quad D) \ 5 \cdot 10^{6} \quad E) \ 5 \cdot 10^{7} \]

Solution:

\[
|z| = \left| \displaystyle \frac{ (-8 \ \ + \ \ 6i)^{5} }{ \sqrt[3]{ 7 \ \ + \ \ \sqrt{15}i } } \right|
= \displaystyle \frac{ | -8 \ \ + \ \ 6i |^{5} }{ \left| 7 \ \ + \ \ \sqrt{15}i \right|^{1/3} }
\]

\[
= \displaystyle \frac{ \left(\sqrt{(-8)^2 + 6^2}\right)^{5} }{ \left(\sqrt{7^2 + (\sqrt{15})^2}\right)^{1/3} } = \frac{ 10^{5} }{ 64^{1/6} } = \frac{ 10^{5} }{ (2^6)^{1/6} } = \frac{10^5}{2} = \frac{10 \cdot 10^4}{2} = 5 \cdot 10^{4}
\]

\( \textbf{Answer: B} \)

QUESTION 19

What is the modulus of the complex number \( z = \displaystyle \frac{ 1 \ \ – \ \ x \ \ + \ \ xi }{ x \ \ – \ \ 1 \ \ – \ \ xi } \)?

\[ A) \ 1 \quad B) \ x \quad C) \ \displaystyle \frac{1}{x} \quad D) \ 2x \quad E) \ \displaystyle \frac{1}{2x} \]

Solution:

\[
|z| = \left| \displaystyle \frac{ 1 \ \ – \ \ x \ \ + \ \ xi }{ x \ \ – \ \ 1 \ \ – \ \ xi } \right| = \displaystyle \frac{ | 1 \ \ – \ \ x \ \ + \ \ xi | }{ | x \ \ – \ \ 1 \ \ – \ \ xi | }
\]

\[
= \displaystyle \frac{ \sqrt{ (1 \ \ – \ \ x)^{2} \ \ + \ \ x^{2} } }{ \sqrt{ (x \ \ – \ \ 1)^{2} \ \ + \ \ (-x)^{2} } }
\]

Since \( (1-x)^2 = (x-1)^2 \) and \( x^2 = (-x)^2 \), the numerator and denominator are identical:

\[
= 1
\]

\( \textbf{Answer: A} \)

QUESTION 20

If \( z_{1}, z_{2} \) are the roots of the quadratic equation:

\[ iz^{2} \ \ – \ \ (\sqrt{3} \ \ – \ \ i)z \ \ + \ \ \sqrt{7} \ \ + \ \ 3i = 0 \]

what is the value of the expression \( \left| \displaystyle \frac{1}{z_{1}} \ \ + \ \ \frac{1}{z_{2}} \right| \)?

\[ A) \ \displaystyle \frac{1}{3} \quad B) \ \displaystyle \frac{1}{2} \quad C) \ 1 \quad D) \ 2 \quad E) \ 3 \]

Solution:

By finding a common denominator, the expression can be rewritten as:

\[ \left| \displaystyle \frac{1}{z_{1}} +\displaystyle \frac{1}{z_{2}} \right| = \left| \displaystyle\frac{ z_{1} + z_{2} }{ z_{1} z_{2} } \right| \]

Using Vieta’s formulas for the sum of roots (\( z_1 + z_2 = -b/a \)) and product of roots (\( z_1 z_2 = c/a \)):

\[ = \left| \displaystyle\frac{ \displaystyle\frac{ \sqrt{3} \ – \ i }{i} } { \displaystyle\frac{ \sqrt{7} + 3i }{i} } \right| = \left| \displaystyle\frac{ \sqrt{3} \ – \ i }{ \sqrt{7} + 3i } \right| \]

Applying the division property of the modulus:

\[ =\displaystyle \frac{ |\sqrt{3} \ – \ i| }{ |\sqrt{7} + 3i| } = \frac{\sqrt{(\sqrt{3})^2 + (-1)^2}}{\sqrt{(\sqrt{7})^2 + 3^2}} = \displaystyle\frac{ 2 }{ \sqrt{16} } = \frac{2}{4} = \frac{1}{2} \]

\( \textbf{Answer: B} \)

QUESTION 21

If \( z_{1}, z_{2} \) are the roots of the equation:

\[
\displaystyle \frac{ z – 1 – \displaystyle \frac{2}{z} }{ z – 2 } + z = 0
\]

what is the value of \( |z_{1}| \)?

\[ A) \ 5 \quad B) \ 4 \quad C) \ 3 \quad D) \ 2 \quad E) \ 1 \]

Solution:

First, find a common denominator in the numerator:

\[ \displaystyle \frac{ z – 1 – \displaystyle \frac{2}{z} }{ z – 2 } + z = 0 \Rightarrow \displaystyle \frac{ \displaystyle\frac{ z^{2} – z – 2 }{z} }{ z – 2 } + z = 0 \]

Factor the quadratic expression \( z^2 – z – 2 = (z – 2)(z + 1) \):

\[
\Rightarrow \displaystyle \frac{ (z – 2)(z + 1) }{ z(z – 2) } + z = 0
\]

Under the restrictions \( z \neq 2 \) and \( z \neq 0 \), we can cancel out \( (z – 2) \):

\[
\Rightarrow \frac{z + 1}{z} + z = 0
\]

\[
\Rightarrow z + 1 + z^{2} = 0 \Rightarrow z^{2} + z + 1 = 0
\]

Using the quadratic formula \( z = \displaystyle\frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \):

\[
\Rightarrow z_{1,2} = \frac{ -1 \ \pm \ \sqrt{ 1^2 – 4(1)(1) } }{ 2 \cdot 1 } = -\frac{1}{2} \ \pm \ \frac{ \sqrt{3} }{ 2 } i
\]

Now, let us calculate the modulus of the roots:

\[
|z_{1}| = |z_{2}| = \sqrt{ \left( -\frac{1}{2} \right)^{2} + \left( \frac{ \sqrt{3} }{ 2 } \right)^{2} } = \sqrt{\frac{1}{4} + \frac{3}{4}} = 1
\]

\( \textbf{Answer: E} \)