We know that a complex number is of the form **z=a+ib** where a and b are real numbers.

Consider two complex numbers z_{1} = a_{1} + ib_{1} and z_{2} = a_{2} + ib_{2}

Then the addition of the complex numbers z_{1} and z_{2} is defined as,

z_{1}+z_{2} =( a_{1}+a_{2} )+i( b_{1}+b_{2} )

We can see that, the real part of the resulting complex number is sum of the real part of the each complex numbers and the imaginary part of the resulting complex number is equal to sum of the imaginary part of the each complex numbers.

That is, Re(z_{1}+z_{2} )= Re( z_{1} )+Re( z_{1} )

Im( z_{1}+z_{2} )=Im( z_{1})+Im(z_{2})

For the complex numbers, z_{1} = a_{1}+ib_{1}

z_{2} = a_{1}+ib_{2}

z_{3} = a_{3}+ib_{3}

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z_{n} = a_{n}+ib_{n}

a_{1}+a_{2}+a_{3}+….+a_{n} = (a_{1}+a_{2}+a_{3}+….+a_{n} )+i(b_{1}+b_{2}+b_{3}+….+b_{n})

Let’s Work Out: Example: z Find the value of a and b if z Solution: By the definition of addition of two complex numbers, Re(z 6 = a + 4 a = 6 – 4 = 2 Im(z 10 = 3+b b = 10-3 =7 |

Note: Conjugate of a complex number z=a+ib is given by changing the sign of the imaginary part of z which is denoted as \( \bar z \)

\( \bar z = a-ib\)

\(z+ \bar z \) =2a

\(z- \bar z \) =2bi

Consider the complex numbers z_{1} = a_{1}+ib_{1} and z_{2} = a_{2}+ib_{2}, then the difference of z_{1} and z_{2}, z_{1}-z_{2} is defined as,

z_{1}-z_{2} = (a_{1}-a_{2})+i(b_{1}-b_{2})

From the definition, it is understood that,

Re(z_{1}-z_{2})=Re(z_{1})-Re(_{2})

Im(z_{1}-z_{2})=Im(z_{1})-ImRe(_{2})

Example: z Solution: By the definition of difference of two complex numbers, Im 0 = a – 4 a = 4 |

Note: All real numbers are complex numbers with imaginary part as zero.

We know the expansion of (a+b)(c+d)=ac+ad+bc+bd

Similarly, consider the complex numbers z_{1} = a_{1}+ib_{1} and z_{2} = a_{2}+ib_{2}

Then, the product of z_{1} and z_{2} is defined as

z_{1} z_{2}=(a_{1}+ib_{1})(a_{2}+ib_{2})

\(z_1 z_2 = a_1 a_2+a_1 b_2 i+b_1 a_2 i+b_1 b_2 i^2\)

Since, \( i^2\)=-1,

\(z_1 z_2 = (a_1 a_2-b_1 b_2 )+i(a_1 b_2+a_2 b_1 )\)

Example: z Solution: z = 6 × 4 + 6 × 3i + (-2i) × 4 + (-2i)(3i) = 24 + 18i – 8i – 6i = 24 + 10i + 6 = 30 + 10i |

Note: Multiplicative inverse of a complex number

Definition: For any non-zero complex number z=a+ib(a≠0 and b≠0) there exists a another complex number \(z^{-1} ~or~ \frac {1}{z}\) which is known as the multiplicative inverse of z such that \(zz^{-1} = 1\).

z = a+ib, then \(z^{-1} = \frac{a}{a^2 + b^2} + i \frac{(-b)}{a^2 + b^2}\)

\(Re(z^{-1}) = \frac{a}{a^2 + b^2}\)

\(Im(z^{-1}) = \frac{-b}{a^2 + b^2}\)

Example: z = 3 + 4i Solution: \(z^{-1}\) of \(a + ib\) = \(\frac{a}{a^2 + b^2} +i \frac{(-b)}{a^2 + b^2}\) = \(\frac{(a-ib)}{a^2 + b^2}\) Numerator of \(z^{-1}\) is conjugate of z, that is a – ib Denominator of \(z^{-1}\) is sum of squares of the Real part and imaginary part of z Here, \(z\) = \( 3 + 4i \) \(z^{-1}\) = \(\frac{3-4i}{3^2 + 4^2}\) = \(\frac{3-4i}{25}\) \(z^{-1}\) = \(\frac{3}{25} – \frac{4i}{25}\) |

Consider the complex number \(z_1\) = \( a_1 + ib_1\) and \(z_2\) =\( a_2 + ib_2\), then the quotient \({z_1}{z_2}\) is defined as,

\(\frac{z_1}{z_2}\) = \(z_1 × \frac{1}{z_2}\)

Therefore, to find \(\frac{z_1}{z_2}\) , we have to multiply \(z_1\) with the multiplicative inverse of \(z_2\).

Example: \(z_1\) = \( 2 + 3i\) and \(z_2\) = \(1 + i\), Find \(\frac{z_1}{z_2}\). Solution: \(\frac{z_1}{z_2}\) = \(z_1 × \frac{1}{z_2}\) \(\frac{2+3i}{1+i} \) = \((2+3i) × \frac{1}{1+i}\) ∵ \( \frac{1}{1+i} \) = \(\frac{1-i}{1^2 + 1^2}\) = \(\frac{1-i}{2}\) \( \frac{2 + 3i}{1 + i} \) = \( 2+3i × \frac{1-i}{2}\)= \( \frac{(2+3i)(1-i)}{2}\) =\(\frac{2 – 2i + 3i – 3i^2}{2} \)= \(\frac{5+i}{2}\) |