Arithmetic Operations on Complex Numbers
1. Addition and Subtraction:
When adding or subtracting complex numbers, the real parts and imaginary parts are added or subtracted separately among themselves.
Examples:
- \( (3 – 3\sqrt{2}i) + (2 + 2\sqrt{2}i) \)
\( = (3 + 2) + (-3\sqrt{2} + 2\sqrt{2})i \)
\( = 5 \ – \ \sqrt{2}i \)
- \( (\sqrt{5} \ – \ \sqrt{3} + \sqrt{2}i) \ – \ (\sqrt{3} \ – \ \sqrt{5} + \sqrt{2}i) \)
\( = \sqrt{5} \ – \ \sqrt{3} + \sqrt{2}i \ – \ \sqrt{3} + \sqrt{5} \ – \ \sqrt{2}i = 2(\sqrt{5} \ – \ \sqrt{3}) \)
- \( (-3 \ – \ \sqrt{-4} ) + ( 4 + \sqrt{-16} ) \)
\( \left( -3 \ – \ \sqrt{-4} \right) + \left( 4 + \sqrt{-16} \right) \)
\( = \left( -3 \ \ – \ \ 2i \right) + \left( 4 \ \ + \ \ 4i \right) \)
\( = 1 + 2i \)
2. Multiplication:
When multiplying complex numbers, the distributive property of multiplication over addition is applied, followed by substituting $i^2 = -1$.
Examples:
- \(\left( 3 + 2i \right)\left( 4 \ \ – \ \ 5i \right)\)
\( = 3\cdot 4 \ \ – \ \ 3\cdot 5i \ \ + \ \ 4\cdot 2i \ \ – \ \ 5i\cdot 2i \)
\( = 12 \ \ – \ \ 15i \ \ + \ \ 8i \ \ + \ \ 10 \)
\( = 22 \ \ – \ \ 7i \)
- \(\left( \sqrt{2} + 2i \right)\left( 2 + \sqrt{2} i \right)\)
\( = 2\sqrt{2} + \sqrt{2}\cdot \sqrt{2} i + 2\cdot 2i + 2i\cdot \sqrt{2} i \)
\( = 2\sqrt{2} + 2i + 4i \ \ – \ \ 2\sqrt{2} \)
\( = 6i \)
- \( \sqrt{-7} \cdot \sqrt{-21} \cdot \sqrt{-3} \)
\( = \sqrt{7}i \cdot \sqrt{21}i \cdot \sqrt{3}i \)
\( = \sqrt{7\cdot 21\cdot 3}\ i^{3} \)
\( = \ -21i \)
3. Division:
When dividing complex numbers, both the numerator and the denominator are multiplied by the complex conjugate of the denominator to rationalize it.
Examples:
- \(\displaystyle \frac{ 2 \ \ – \ \ i }{ 3 \ \ – \ \ 2i }\)
\( = \displaystyle \frac{ \left( 2 \ \ – \ \ i \right)\left( 3 \ \ + \ \ 2i \right) }{ \left( 3 \ \ – \ \ 2i \right)\left( 3 \ \ + \ \ 2i \right) } \)
\( = \displaystyle \frac{ 2\cdot 3 + 2\cdot 2i \ \ – \ \ 3i \ \ – \ \ i\cdot 2i }{ 3^{2} + \left( -2 \right)^{2} } \)
\( = \displaystyle \frac{ 8 + i }{ 13 } \)
- \( \displaystyle \frac{ \sqrt{3} \ \ – \ \ \sqrt{2} i }{ \sqrt{2} \ \ – \ \ \sqrt{3} i } \)
\( = \displaystyle \frac{ \left( \sqrt{3} \ \ – \ \ \sqrt{2} i \right)\left( \sqrt{2} \ \ + \ \ \sqrt{3} i \right) }{ \left( \sqrt{2} \ \ – \ \ \sqrt{3} i \right)\left( \sqrt{2} \ \ + \ \ \sqrt{3} i \right) } \)
\( = \displaystyle \frac{ \sqrt{6} + 3i \ \ – \ \ 2i + \sqrt{6} }{ (\sqrt{2})^{2} + (\sqrt{3})^{2} } \)
\( = \displaystyle \frac{ 2\sqrt{6} + i }{ 5 } \)
- \( \displaystyle \frac{1}{1 \ \ + \ \ (1 \ \ + \ \ \sqrt{2})i} \)
\( \displaystyle \frac{1 \cdot \left( 1 \ \ – \ \ (1 \ \ + \ \ \sqrt{2})i \right)}{ \left( 1 \ \ + \ \ (1 \ \ + \ \ \sqrt{2})i \right)\left( 1 \ \ – \ \ (1 \ \ + \ \ \sqrt{2})i \right) } \)
\( \displaystyle \frac{ 1 \ \ – \ \ (1 \ \ + \ \ \sqrt{2})i }{ 1^{2} \ \ + \ \ (1 \ \ + \ \ \sqrt{2})^{2} } \)
\( \displaystyle \frac{ 1 \ \ – \ \ (1 \ \ + \ \ \sqrt{2})i }{ 4 \ \ + \ \ 2\sqrt{2} } \)
QUESTION 8
Given the equation:
\[ z \ \ – \ \ \sqrt{2} = (\sqrt{2}z \ \ + \ \ 1)i \]
what is the imaginary part of the complex number \( z \) ?
\[ A) \ 1 \quad B) \ 0 \quad C) \ 1 \quad D) \ 2 \quad E) \ 3 \]
Solution:
\( z \ \ – \ \ \sqrt{2} = (\sqrt{2}z \ \ + \ \ 1)i \Rightarrow z \ \ – \ \ \sqrt{2} = \sqrt{2}z i \ \ + \ \ i \)
\( \Rightarrow z \ \ – \ \ \sqrt{2}z i = \sqrt{2} \ \ + \ \ i \)
\( \Rightarrow z(1 \ \ – \ \ \sqrt{2}i) = \sqrt{2} \ \ + \ \ i \)
\( \Rightarrow z = \displaystyle \frac{ \sqrt{2} \ \ + \ \ i }{ 1 \ \ – \ \ \sqrt{2}i } \)
\( \Rightarrow z = \displaystyle \frac{ (\sqrt{2} \ \ + \ \ i)(1 \ \ + \ \ \sqrt{2}i) }{ (1 \ \ – \ \ \sqrt{2}i)(1 \ \ + \ \ \sqrt{2}i) } \)
\( \Rightarrow z = \displaystyle \frac{ \sqrt{2} \ \ + \ \ 2i \ \ + \ \ i \ \ – \ \ \sqrt{2} }{ 1^{2} \ \ + \ \ (-\sqrt{2})^{2} } = \displaystyle \frac{ 3i }{ 3 } = i \)
Since $z = i$, we find that \( \mathrm{Im}(z) = 1 \)
\( \textbf{Answer: C} \)
QUESTION 9
Given that \( \mathrm{Im}(z) \neq 0 \), which of the following choices represents the complex number \( z \) that satisfies the equation \( z^{2} \ \ – \ \ 2\bar{z} \ \ – \ \ 2 = 0 \) ?
\[ A) \ 1 + i \quad B) \ -1 + i \quad C) \ 2 + i \quad D) \ 2 \ – \ i \quad E) \ -2 \ – \ i \]
Solution:
Let \( z = a \ \ + \ \ bi \), which implies \( \bar{z} = a \ \ – \ \ bi \).
\( z^{2} \ \ – \ \ 2\bar{z} \ \ – \ \ 2 = 0 \)
\( \Rightarrow (a \ \ + \ \ bi)^{2} \ \ – \ \ 2(a \ \ – \ \ bi) \ \ – \ \ 2 = 0 \)
\( \Rightarrow a^{2} \ \ – \ \ b^{2} \ \ + \ \ 2(ab)i \ \ – \ \ 2a \ \ + \ \ 2bi \ \ – \ \ 2 = 0 \)
\( \Rightarrow ( a^{2} \ \ – \ \ b^{2} \ \ – \ \ 2a \ \ – \ \ 2 ) \ \ + \ \ (2ab \ \ + \ \ 2b)i = 0 \)
Setting both the real and imaginary parts to 0:
\( a^{2} \ \ – \ \ b^{2} \ \ – \ \ 2a \ \ – \ \ 2 = 0 \) and \( 2ab \ \ + \ \ 2b = 0 \)
\( 2b(a \ \ + \ \ 1) = 0 \)
Since \( \mathrm{Im}(z) = b \neq 0 \), we must have \( a = -1 \). Substituting \( a = -1 \) into the real part equation yields:
\( \Rightarrow (-1)^{2} \ \ – \ \ b^{2} \ \ – \ \ 2\cdot(-1) \ \ – \ \ 2 = 0 \)
\( \Rightarrow 1 \ \ – \ \ b^{2} \ \ + \ \ 2 \ \ – \ \ 2 = 0 \Rightarrow b^2 = 1 \)
\( \Rightarrow b = 1 \) (or $b = -1$, but looking at the options, $b = 1$ gives \( z = -1 \ \ + \ \ i \))
\( \textbf{Answer: B} \)
QUESTION 10
If
\[ \displaystyle \frac{ 2z\bar{z} }{ z \ \ + \ \ \bar{z} } = i(\bar{z} \ \ – \ \ z) \]
what is the value of the difference \( \mathrm{Re}(z) \ – \ \mathrm{Im}(z) \) ?
\[ A) \ 0 \quad B) \ 1 \quad C) \ 2 \quad D) \ 3 \quad E) \ 4 \]
Solution:
Let \( z = a \ \ + \ \ bi \), then \( \bar{z} = a \ \ – \ \ bi \).
\( \displaystyle \frac{ 2z\bar{z} }{ z \ \ + \ \ \bar{z} } = i(\bar{z} \ \ – \ \ z) \Rightarrow \frac{ 2(a^{2} \ \ + \ \ b^{2}) }{ a \ \ + \ \ bi \ \ + \ \ a \ \ – \ \ bi } = i(a \ \ – \ \ bi \ \ – \ \ (a \ \ + \ \ bi)) \)
\( \Rightarrow \displaystyle \frac{ 2(a^{2} \ \ + \ \ b^{2}) }{ 2a } = i(-2bi) \)
\( \Rightarrow \displaystyle \frac{ a^{2} \ \ + \ \ b^{2} }{ a } = 2b \)
\( \Rightarrow a^{2} \ \ + \ \ b^{2} \ \ – \ \ 2ab = 0 \)
\( \Rightarrow (a \ \ – \ \ b)^{2} = 0 \Rightarrow a \ \ – \ \ b = 0 \)
Therefore, \( \mathrm{Re}(z) \ – \ \mathrm{Im}(z) = a – b = 0 \).
\( \textbf{Answer: A} \)
QUESTION 11
Given the polynomial \( P(x) = x^{4} \ \ – \ \ 3x^{3} \ \ + \ \ 4x^{2} \ \ + \ \ 2x \), what is the value of \( P(1 \ \ + \ \ i) \) ?
\[ A) \ 1+i \quad B) \ 2+2i \quad C) \ 3+3i \quad D) \ 4+4i \quad E) \ 5+5i \]
Solution:
\( P(x) = x^{4} \ \ – \ \ 3x^{3} \ \ + \ \ 4x^{2} \ \ + \ \ 2x \)
\( \Rightarrow P(x) = x(x^{3} \ \ – \ \ 3x^{2} \ \ + \ \ 4x \ \ + \ \ 2) \)
\( \Rightarrow P(x) = x(x^{3} \ \ – \ \ 3x^{2} \ \ + \ \ 3x \ \ – \ \ 1 \ \ + \ \ x \ \ + \ \ 3) \)
\( \Rightarrow P(x) = x[(x \ \ – \ \ 1)^{3} \ \ + \ \ x \ \ + \ \ 3] \)
Now, let us evaluate \( P(1 \ \ + \ \ i) \):
\( \Rightarrow P(1 \ \ + \ \ i) = (1 \ \ + \ \ i)\left((1 \ \ + \ \ i \ \ – \ \ 1)^{3} \ \ + \ \ 1 \ \ + \ \ i \ \ + \ \ 3\right) \)
\( \Rightarrow P(1 \ \ + \ \ i) = (1 \ \ + \ \ i)(i^{3} \ \ + \ \ 4 \ \ + \ \ i) \)
\( \Rightarrow P(1 \ \ + \ \ i) = (1 \ \ + \ \ i)(-i \ \ + \ \ 4 \ \ + \ \ i) \)
\( \Rightarrow P(1 \ \ + \ \ i) = (1 \ \ + \ \ i)(4) = 4 \ \ + \ \ 4i \)
\( \textbf{Answer: D} \)
QUESTION 12
What is the result of the following expression?
\[ \displaystyle \frac{ \sqrt{-4} \cdot \sqrt{-9} }{ \sqrt{-2} \cdot \sqrt{-3} \cdot \sqrt{-6} } \]
\[ A) \ 1 \quad B) \ – 1 \quad C) \ i \quad D) \ -i \quad E) \ 2 \]
Solution:
\[ \displaystyle \frac{ \sqrt{-4} \cdot \sqrt{-9} }{ \sqrt{-2} \cdot \sqrt{-3} \cdot \sqrt{-6} } = \frac{ 2i \cdot 3i }{ \sqrt{2}i \cdot \sqrt{3}i \cdot \sqrt{6}i } \]
\[ = \displaystyle \frac{ -6 }{ \sqrt{2\cdot 3\cdot 6}\ i^{3} } \]
\[ = \displaystyle \frac{ -6 }{ -6i } = \frac{1}{i} = \frac{ -i }{ i\cdot(-i) } = -i \]
\( \textbf{Answer: D} \)
QUESTION 13
What is the imaginary part of the complex number \( z = (\sqrt{2} \ \ + \ \ i)^{5} \cdot (\sqrt{2} \ \ – \ \ i)^{6} \) ?
\[ A) \ -81 \quad B) \ 81 \quad C) \ -81 \sqrt{ 2} \quad D) \ 243 \quad E) \ -243 \]
Solution:
\[ z = (\sqrt{2} \ \ + \ \ i)^{5} \cdot (\sqrt{2} \ \ – \ \ i)^{6} \]
\[ = (\sqrt{2} \ \ + \ \ i)^{5} \cdot (\sqrt{2} \ \ – \ \ i)^{5} \cdot (\sqrt{2} \ \ – \ \ i) \]
\[ = [(\sqrt{2} \ \ + \ \ i)\cdot(\sqrt{2} \ \ – \ \ i)]^{5} \cdot (\sqrt{2} \ \ – \ \ i) \]
\[ = [(\sqrt{2})^{2} \ \ + \ \ 1^{2}]^{5} \cdot (\sqrt{2} \ \ – \ \ i) \]
\[ = 3^{5} \cdot (\sqrt{2} \ \ – \ \ i) = 243(\sqrt{2} \ \ – \ \ i) = 243\sqrt{2} \ \ – \ \ 243i \]
Therefore, \( \mathrm{Im}(z) = -243 \).
\( \textbf{Answer: E} \)
QUESTION 14
What is the imaginary part of the complex number \( z = \displaystyle \frac{1}{\sqrt{2} \ \ – \ \ 1 \ \ + \ \ i} \ \ – \ \ \frac{1}{\sqrt{2} \ \ – \ \ 1 \ \ – \ \ i} \ \ + \ \ \frac{i}{\sqrt{2}} \)?
\[ A) \ -1 \quad B) \ 1 \quad C) \ -2 \quad D) \ 2 \quad E) \ 0 \]
Solution:
\[ z = \displaystyle \frac{1}{\sqrt{2} \ \ – \ \ 1 \ \ + \ \ i} \ \ – \ \ \frac{1}{\sqrt{2} \ \ – \ \ 1 \ \ – \ \ i} \ \ + \ \ \frac{i}{\sqrt{2}} \]
\[ = \displaystyle \frac{ \sqrt{2} \ \ – \ \ 1 \ \ – \ \ i \ \ – \ \ (\sqrt{2} \ \ – \ \ 1 \ \ + \ \ i) }{ (\sqrt{2} \ \ – \ \ 1 \ \ + \ \ i)(\sqrt{2} \ \ – \ \ 1 \ \ – \ \ i) } \ \ + \ \ \frac{i}{\sqrt{2}} \]
\[ = \displaystyle \frac{ -2i }{ (\sqrt{2} \ \ – \ \ 1)^{2} \ \ + \ \ 1^{2} } \ \ + \ \ \frac{i}{\sqrt{2}} \]
\[ = \displaystyle \frac{ -2i }{ 2 – 2\sqrt{2} + 1 + 1 } \ \ + \ \ \frac{i}{\sqrt{2}} = \frac{ -2i }{ 4 \ \ – \ \ 2\sqrt{2} } \ \ + \ \ \frac{i}{\sqrt{2}} \]
\[ = \displaystyle \frac{ -2i \cdot (4 \ \ + \ \ 2\sqrt{2}) }{ 4^2 – (2\sqrt{2})^2 } \ \ + \ \ \frac{i}{\sqrt{2}} = \frac{ -8i \ \ – \ \ 4\sqrt{2}i }{ 16 – 8 } \ \ + \ \ \frac{i}{\sqrt{2}} \]
\[ = \displaystyle \frac{ -8i \ \ – \ \ 4\sqrt{2}i }{ 8 } \ \ + \ \ \frac{ \sqrt{2}i }{ 2 } \]
\[ = -i \ \ – \ \ \displaystyle\frac{\sqrt{2}i}{2} \ \ + \ \ \displaystyle\frac{ \sqrt{2}i }{ 2 } = -i \]
Hence, \( \mathrm{Im}(z) = -1 \)
\( \textbf{Answer: A} \)
Tip:
For any integer \( n \in \mathbb{Z} \):
\[ (a \ \ \pm \ \ ai)^{2n} = [(a \ \ \pm \ \ ai)^{2}]^{n} = (\pm 2a^{2} i)^{n} \]
Example:
Find the result of the operation \( (2 \ \ + \ \ 2i)^{22} \).
\( (2 \ \ + \ \ 2i)^{22} = [(2 \ \ + \ \ 2i)^{2}]^{11} = (4 \ \ + \ \ 8i \ \ – \ \ 4)^{11} \)
\( = (8i)^{11} = 2^{33} i^{11} = 2^{33} (-i) = -2^{33} i \)
Example:
Find the result of the operation \( (\sqrt{2} \ \ – \ \ \sqrt{2}i)^{33} \).
\( (\sqrt{2} \ \ – \ \ \sqrt{2}i)^{33} = (\sqrt{2} \ \ – \ \ \sqrt{2}i)^{32} \cdot (\sqrt{2} \ \ – \ \ \sqrt{2}i) \)
\( = [(\sqrt{2} \ \ – \ \ \sqrt{2}i)^{2}]^{16} \cdot (\sqrt{2} \ \ – \ \ \sqrt{2}i) \)
\( = (2 \ \ – \ \ 4i \ \ – \ \ 2)^{16} \cdot (\sqrt{2} \ \ – \ \ \sqrt{2}i) \)
\( = (-4i)^{16} \cdot (\sqrt{2} \ \ – \ \ \sqrt{2}i) \)
\( = 4^{16} \cdot i^{16} \cdot (\sqrt{2} \ \ – \ \ \sqrt{2}i) = 2^{32} \cdot 1 \cdot (\sqrt{2} \ \ – \ \ \sqrt{2}i) = 2^{32}(\sqrt{2} \ \ – \ \ \sqrt{2}i) \)
QUESTION 15
If \( z = \displaystyle \frac{\sqrt{3}}{i} \ \ + \ \ \frac{3}{\sqrt{3}} \), which of the following is equal to \( z^{20} \)?
\[ A) \ 3^{20} \quad B) \ 6^{10} \quad C) \ -6^{10} \quad D) \ -3^{20} \quad E) \ 3^{10} \]
Solution:
\[ z = \displaystyle \frac{\sqrt{3}}{i} \ \ + \ \ \frac{3}{\sqrt{3}} \Rightarrow z = – \sqrt{3}i \ \ + \ \ \sqrt{3} = \sqrt{3}(1 – i) \]
\[ \Rightarrow z^{20} = \left( \sqrt{3}(1 – i) \right)^{20} = (\sqrt{3})^{20} \cdot (1 – i)^{20} \]
\[ = 3^{10} \cdot \left[ (1 – i)^2 \right]^{10} \]
\[ = 3^{10} \cdot (-2i)^{10} = 3^{10} \cdot 2^{10} \cdot i^{10} \]
\[ = 6^{10} \cdot (-1) = -6^{10} \]
\( \textbf{Answer: C} \)
QUESTION 16
If \( (1 \ \ + \ \ i)z \ \ + \ \ i = 1 \), which of the following is equal to \( z^{1996} \) ?
\[ A) \ -i \quad B) \ i \quad C) \ -1 \quad D) \ 0 \quad E) \ 1 \]
Solution:
\[ (1 \ \ + \ \ i)z \ \ + \ \ i = 1 \Rightarrow (1 \ \ + \ \ i)z = 1 \ \ – \ \ i \]
\[ \Rightarrow z = \displaystyle \frac{ 1 \ \ – \ \ i }{ 1 \ \ + \ \ i } = \frac{(1-i)^2}{1^2+1^2} = \frac{-2i}{2} = -i \]
\[ \Rightarrow z^{1996} = (-i)^{1996} = i^{1996} \]
Since \( 1996 \equiv 0 \pmod{4} \), \( i^{1996} = i^0 = 1 \)
\( \textbf{Answer: E} \)
QUESTION 17
If \( z = \displaystyle \frac{ 2i }{ 1 \ \ – \ \ i } \), what is the value of \( z^{50} \) ?
\[ A) \ 1 \quad B) \ -2^{25} \quad C) \ 2^{25} \quad D) \ -2^{25}i \quad E) \ 2^{25}i \]
Solution:
\[ z = \displaystyle \frac{ 2i }{ 1 \ \ – \ \ i } = \frac{2i(1+i)}{1^2+(-1)^2} = \frac{2i – 2}{2} = -1 + i \]
\[ \Rightarrow z^{50} = (-1 + i)^{50} = \left[ (-1 + i)^2 \right]^{25} \]
\[ = (-2i)^{25} = (-2)^{25} \cdot i^{25} \]
\[ = -2^{25} \cdot i^1 = -2^{25}i \]
\( \textbf{Answer: D} \)