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Real And Complex Numbers
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Arithmetic Operations on Complex Numbers

Real And Complex Numbers

Arithmetic Operations on Complex Numbers

Math

  • Addition

    To add two complex numbers, we add their real and imaginary parts separately. That is, if we have two complex numbers z1=a+biz_1 = a + bi and z2=c+diz_2 = c + di, then their sum is given by:

    z1+z2=(a+c)+(b+d)iz_1 + z_2 = (a + c) + (b + d)i

  • Subtraction

    To subtract two complex numbers, we subtract their real and imaginary parts separately. That is, if we have two complex numbers z1=a+biz_1 = a + bi and z2=c+diz_2 = c + di, then their difference is given by:

    z1−z2=(a−c)+(b−d)iz_1 - z_2 = (a - c) + (b - d)i

  • Multiplication

    To multiply two complex numbers, we use the distributive law of multiplication over addition and the fact that i2=1i^2=1. That is, if we have two complex numbers z1=a+biz_1 = a + bi and z2=c+diz_2 = c + di, then their product is given by: (a+bi)(c+di)=(ac−bd)+(ad+bc)i(a + bi)(c + di) = (ac - bd) + (ad + bc)i

  • Division

    Division of complex numbers is performed by multiplying both the numerator and denominator by the complex conjugate of the denominator. This allows us to get rid of the complex number in the denominator and express the result as a single complex number. For two complex numbersz1=a+biz_1 = a + bi and z2=c+diz_2 = c + di, their division is given by:

    (a+bi)(c+di)=(ac+bd)(c2+d2)+(bc−ad)(c2+d2)i\frac{(a + bi) }{ (c + di)} = \frac{(ac + bd)} { (c^2 + d^2) }+ \frac{(bc - ad) }{ (c^2 + d^2)i}

Examples

  • Addition

    Let's consider an example to illustrate addition of complex numbers. Suppose we have two complex numbers:

    z1=3+4iz2=−2+6iz_1 = 3 + 4i\\ z_2 = -2 + 6i

    we know that, <strong>z1+z2=(a+c)+(b+d)i</strong><strong>z_1 + z_2 = (a + c) + (b + d)i</strong>

    To find their sum, we add their real and imaginary parts separately:

    z1+z2=(3+(−2))+(4+6)iz1+z2=1+10iz_1 + z_2 = (3 + (-2)) + (4 + 6)i \\z_1 + z_2= 1 + 10i

    Therefore, the sum of the complex numbers z1z_1 and z2z_2 is 1+10i1 + 10i.

    Solution: Given

    z1=3+4iz2=−2+6iz_1 = 3 + 4i\\ z_2 = -2 + 6i

    Formula z1+z2=(a+c)+(b+d)iz_1 + z_2 = (a + c) + (b + d)i

    z1+z2=(3+(−2))+(4+6)iz1+z2=1+10iz_1 + z_2 = (3 + (-2)) + (4 + 6)i \\z_1 + z_2= 1 + 10i

  • Subtraction

    Let's consider an example to illustrate subtraction of complex numbers. Suppose we have two complex numbers:

    z1=3+4iz2=−2+6iz_1 = 3 + 4i\\ z_2 = -2 + 6i

    To find their difference, we subtract their real and imaginary parts separately:

    z1−z2=(3−(−2))+(4−6)iz1−z2=5−2iz_1 - z_2 = (3 - (-2)) + (4 - 6)i \\z_1 - z_2 = 5 - 2i

    Therefore, the difference of the complex numbers z1z_1 and z2z_2 is 5−2i5 - 2i.

    **Solution

    Given**

    z1=3+4iz2=−2+6iz_1 = 3 + 4i\\ z_2 = -2 + 6i

    Formula z1−z2=(a−c)+(b−d)iz_1 - z_2 = (a - c) + (b - d)i

    z1−z2=(3−(−2))+(4−6)iz1−z2=5−2iz_1 - z_2 = (3 - (-2)) + (4 - 6)i \\z_1 - z_2 = 5 - 2i

  • Multiplication

    Let's consider an example to illustrate multiplication of complex numbers. Suppose we have two complex numbers:

    z1=3+4iz2=−2+6iz_1 = 3 + 4i \\z_2 = -2 + 6i

    To find their product, we use the formula (a+bi)(c+di)=(ac−bd)+(ad+bc)i(a + bi)(c + di) = (ac - bd) + (ad + bc)i

    and simplify:

    z1<em>z2=(3</em>(−2)−4<em>6)+(3</em>6+4<em>(−2))iz1</em>z2=(−6−24)+(18−8)iz1∗z2=(−30+10i)z_1 <em> z_2 = (3 </em> (-2) - 4 <em> 6) + (3 </em> 6 + 4 <em> (-2))i\\z_1 </em> z_2 =(-6-24)+(18-8)i\\ z_1 * z_2 = (-30 + 10i)

    Therefore, the product of the complex numbers z1z_1 and z2z_2 is −30+10i-30 + 10i.

    Solution:

    Given z1=3+4iz2=−2+6iz_1 = 3 + 4i \\z_2 = -2 + 6i

    Formula (a+bi)(c+di)=(ac−bd)+(ad+bc)i(a + bi)(c + di) = (ac - bd) + (ad + bc)i

    z1<em>z2=(3</em>(−2)−4<em>6)+(3</em>6+4<em>(−2))iz1</em>z2=(−6−24)+(18−8)iz1∗z2=(−30+10i)z_1 <em> z_2 = (3 </em> (-2) - 4 <em> 6) + (3 </em> 6 + 4 <em> (-2))i\\z_1 </em> z_2 =(-6-24)+(18-8)i\\ z_1 * z_2 = (-30 + 10i)

  • Division

    Let's take an example to understand division better:

    Example: Divide z1=2+5iz_1 = 2 + 5i by z2=1−3iz_2 = 1 - 3i

    Solution:

    Formula: (a+bi)(c+di)=(ac+bd)(c2+d2)+(bc−ad)(c2+d2)i\frac{(a + bi) }{ (c + di)} = \frac{(ac + bd)} { (c^2 + d^2) }+ \frac{(bc - ad) }{ (c^2 + d^2)i} Step 1: Find the conjugate of the denominator (z2)(z_2). Conjugate of z2=1+3iz_2 = 1 + 3i

    Step 2: Multiply both the numerator and denominator by the conjugate of z2z_2.z1z2=(2+5i)(1+3i)(12−(3i)2)\frac{z_1}{z_2}= \frac{(2 + 5i)(1 + 3i)}{ (1^2 -(3i)^2)}\\Solving Numerator:

    (2+5ι)(1+3ι)=2⋅1+2⋅3ι+5ι⋅1+5ι⋅3ι=2+6ι+5ι+15ι2(2+5\iota)(1+3\iota)=2⋅1+2⋅3\iota+5\iota⋅1+5\iota⋅3\iota=2+6\iota+5\iota+15\iota^2

    Since ι2=−1\iota^2=−1, we can simplify this to: 2+6i+5i−152+6i+5i−15

    Now, combine like terms: −13+11i−13+11i

    Denominator: (1−3i)(1+3i)=1⋅1+1⋅3i−3i⋅1−3i⋅3i=1+3i−3i−9i2(1−3i)(1+3i)=1⋅1+1⋅3i−3i⋅1−3i⋅3i=1+3i−3i−9i^2

    Again, since ι2=−1\iota^2=−1, we can simplify this to: 1+3i−3i+91+3i−3i+9

    Now, combine like terms: 10

    So, the division becomes:

    z1z2=(2+5i+6i−15)(1+9)z1z2=(−13+11i)10\frac{z_1}{z_2}= \frac{(2 + 5i + 6i - 15)} { (1 + 9)}\\ \frac{z_1}{z_2}= \frac{(-13 + 11i) }{ 10}

    Therefore, z1z2=(−13+11i)10\frac{z_1}{z_2}= \frac{(-13 + 11i) }{ 10}