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Last modified on February 22nd, 2024

To multiply two or more complex numbers, we use the distributive property. It is done using the FOIL method, which is also used to multiply two binomials.

On multiplying two complex numbers: z_{1} = a_{1} + ib_{1} and z_{2} = a_{2} + ib_{2}, the product obtained is written as:

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

The general formula to calculate the product of two complex numbers is:

**(a**_{1}** + ib**_{1}**)(a**_{2}** + ib**_{2}**) = (a**_{1}**a**_{2}** – b**_{1}**b**_{2}**) + i(a**_{1}**b**_{2}** + a**_{2}**b**_{1}**)**

Let us now multiply two complex numbers: (1 + 2i) and (3 + 5i)

The product is (1 + 2i)(3 + 5i)

Using the FOIL method, we get 1 âœ• 3 + 1 âœ• 5i + 2i âœ• 3 + 2i âœ• 5i

= 3 + 5i + 6i + 10i^{2}

Replacing the value of i^{2} = -1,

3 + 5i + 6i – 10

Adding the like terms, we get (3 – 10) + (5i + 6i)

= -7 + 11i

Instead of multiplying further likewise, by applying the above formula, we can easily obtain the product as:

(1 + 2i)(3 + 5i)

= (1 âœ• 3 – 2 âœ• 5) + i(1 âœ• 5 + 2 âœ• 3)

= -7 + 11i

The numbers 1 + 2i, 3 + 5i, and their product -7 + 11i give the image in the argand plane, as shown.

Similarly, we can multiply the complex numbers represented in the polar form.

The multiplication of complex numbers z_{1} = r_{1}(cosÎ¸_{1 }+ isinÎ¸_{1}) and z_{2} = r_{2}(cosÎ¸_{2 }+ isinÎ¸_{2}) is given by the formula:

**z**_{1}**z**_{2}** = r**_{1}**r**_{2}**[cos(Î¸**_{1}** + Î¸**_{2}**) + isin(Î¸**_{1}** + Î¸**_{2}**)]**

â‡’ **z**_{1}**z**_{2}** = r(cosÎ¸ + isinÎ¸), here r = r**_{1}**r**_{2}** and Î¸ = (Î¸**_{1}** + Î¸**_{2}**).**

Now, let us multiply the complex numbers z_{1} = 6(cos190Â° + isin190Â°) and z_{2} = 5(cos220Â° + isin220Â°) using the above formula..

On multiplying z_{1} and z_{2}, we get,

z_{1}z_{2} = 6(cos190Â° + isin190Â°) âœ• 5(cos220Â° + isin220Â°)

= 6 âœ• 5 [(cos190Â°cos220Â° – sin190Â°sin220Â°) + i(sin190Â°cos220Â° + cos190Â°sin220Â°)]

= 30[cos(190Â° + 220Â°) + isin(190Â° + 220Â°)]

= 30(cos410Â° + isin410Â°)

= r(cosÎ¸ + isinÎ¸), where r = 6 âœ• 5 and Î¸ = (190Â° + 220Â°) = 410Â°.

The formula to multiply a complex number z = a + ib with a purely real number c is:

c âœ• (a + ib) = ac + ibc

Similarly, the formula to multiply a complex number z = a + ib with a purely imaginary number id is:

(id) âœ• (a + ib) = iad – bd.

Now, multiplying 5 + 4i with the real number 8, we get,

8 âœ• (5 + 4i) = 8 âœ• 5 + 8 âœ• 4i = 40 + 32i.

Again, multiplying 5 + 4i with the imaginary number -2i, we get,

(-2i) âœ• (5 + 4i) = (-2i) âœ• 5 + (-2i) âœ• 4i = -10i – 8i^{2}

By substituting the value of i^{2} = -1, the product is -10i + 8.

Like the real numbers, the complex numbers follow all the properties of multiplication:

- Closure Property: On multiplying two complex numbers, the product is always a complex number. Thus, for z
_{1}and z_{2}, the product is ${z_{1}\cdot z_{2}}$ = (a_{1}a_{2}– b_{1}b_{2}) + i(a_{1}b_{2}+ a_{2}b_{1}), which is a complex number. - Commutative Property: For two complex numbers z
_{1}, and z_{2}, ${z_{1}\cdot z_{2}=z_{2}\cdot z_{1}}$ - Associative Law: For the given three complex numbers z
_{1}, z_{2}, and z_{3}, ${\left( z_{1}\cdot z_{2}\right) \cdot z_{3}}$ = ${z_{1}\cdot \left( z_{2}\cdot z_{3}\right)}$ - Multiplicative Identity: For any complex number z, there exists 1 (1 + 0i), such that ${z\cdot 1=1\cdot z=z}$
- Multiplicative Inverse: The multiplicative inverse of any complex number z is denoted by z
^{-1}and, multiplying with z, gives the product 1, z â‹… z^{-1}= z^{-1}â‹… z = 1.

Here, ${z^{-1}=\dfrac{1}{a+ib}}$ = ${\dfrac{1}{a+ib}\times \dfrac{a-ib}{a-ib}}$ = ${\dfrac{a-ib}{a^{2}+b^{2}}}$ = ${\dfrac{\overline{z}}{\left| z\right| ^{2}}}$

**Multiply (3 + 8i) and (9 + 4i).**

Solution:

As we know, the formula for multiplying two complex numbers is:

(a_{1} + ib_{1})(a_{2} + ib_{2}) = (a_{1}a_{2} – b_{1}b_{2}) + i(a_{1}b_{2} + a_{2}b_{1})

Here, (3 + 8i)(9 + 4i)

= (3 âœ• 9 – 8 âœ• 4) + i(3 âœ• 4 + 8 âœ• 9)

= -5 + 84i

**Find the square of -2 + 4i.**

Solution:

As we know, the formula for multiplying two complex numbers is:

(a_{1} + ib_{1})(a_{2} + ib_{2}) = (a_{1}a_{2} – b_{1}b_{2}) + i(a_{1}b_{2} + a_{2}b_{1})

Here, (-2 + 4i)^{2} = (-2 + 4i)(-2 + 4i)

= [(-2) âœ• (-2) – 4 âœ• 4] + i[(-2) âœ• 4 + 4 âœ• (-2)]

= (4 – 16) + i(-8 – 8)

= -12 – 16i

**Find the multiplicative inverse of 5 + i.**

Solution:

As we know, the multiplicative inverse of any complex number z is

z^{-1} = ${\dfrac{\overline{z}}{\left| z\right| ^{2}}}$

Here, |z| = ${\sqrt{5^{2}+1^{2}}}$ = ${\sqrt{26}}$

â‡’ |z|^{2} = 26

${\overline{z}}$ = ${\overline{5+i}}$ = 5 – i

Thus, z^{-1} = ${\dfrac{5-i}{26}}$

Last modified on February 22nd, 2024