# Operations with complex numbers in binomic form

As we add up and substract real numbers: $$2 +2 = 4$$, we have to learn how these operations work when we deal with complex numbers in their binomic form.

Let's see then how to proceed if for example we want to find the complex number corresponding to the solution of: $$(6-5i)+(3+9i)$$$To add two complex numbers in their binomic form, we simply have to add up each of its components separately. Or in other words, we add the real parts on the one hand and we add the imaginary parts on the other hand. In our example the real parts are $$6$$ and $$3$$, so they add up to $$9$$. The imaginary parts are $$-5i$$ and $$9i$$ so they add up to $$4i$$. Then, the result in binomic form will be a complex number which real part is the sum of the real components and the imaginary part is the sum of the imaginary components. Thats is $$9 +4 i$$. In general, given two complex numbers, which in general can be expressed as follows: $$z_1 =a+bi z_2 =a'+b'i$$$

Their sum is defined as: $$z_1+z_2=(a+bi)+(ab'i)=(a+a')+(b+b')i$$$For example, $$z_1=3+5i , \, z_2=4+7i$$ $$z_1 +z_2 =(3+5i)+(4+7i)=(3+4)+(5+7)i=7+12i$$ or: $$z_1=13+i , \, z_2=4-87i$$ $$z_1 +z_2 =(13+i)+(4-87i)=(13+4)+(1-87)i=17-86i$$ In the case where we want to substract two complex numbers, the process will be the same but, instead of adding the real parts and the imaginary parts separately, now we will also substract them separately. So we carry out the computations of the imaginary and real parts separately, and then we join the results in the complex number $$a+ib$$ where $$a$$ is the result of doing the subtraction of the real parts, and $$b$$ is the result of doing the subtraction of the complex parts. If we have $$(4+21i)-(7-3i)$$$ In this case, the substraction of the real parts is: $$(4-7)$$

and the substraction of the imaginary parts is: $$21i-(-3i)=(21+3)$$

Therefore, joining $$a + bi$$ we obtain: $$(4+21i)-(7-3i)=(4-7)+(21-(-3))=-3+24i$$$or: $$\Big( \Big( \dfrac{3}{8} \Big) +7i\Big) -(\sqrt{2}+5i)= \Big( \dfrac{3}{8} +\sqrt{2} \Big) +(7-5)i= \dfrac{3-8\sqrt{2}}{8}+2i$$$

Let's see how to proceed to solve products of complex numbers expressed in binomic form. We will do it with an example to better understand what we must do.

Let's look for the solution of the product: $$(3+9i)\cdot(4+2i)$$$What we will do is to apply the distributive property of the product with respect to the sum. This is: $$\displaystyle \begin{array}{rl} (3+9i)\cdot(4+2i)&=3\cdot4+3\cdot2i+9i\cdot4+9i\cdot2i \\ & =12+6i+36i+18i^2=12+42i+18i^2 \end{array}$$$

Now, bearing in mind that $$i=\sqrt{-1} \Rightarrow i^2=-1$$ and substituting in the previous expression, we have: $$\displaystyle \begin{array}{rl} (3+9i)\cdot(4+2i) &=12+42i+18i^2=12+42i+18(-1) \\ &=12-18+42i=-6+42i \end{array}$$$Therefore, the real part of the product of two complexes also includes the factor that multiplies $$i^2$$, since it is $$-1$$. In general, we will have: To calculate the product of any two complex numbers that we will denote typically as $$z_1 =a+bi$$$

$$z_2 =a'+b'i$$$we will have to apply the following formula that comes out when applying the distributive property and the property that $$i=\sqrt{-1} \Rightarrow i^2=-1$$. Therefore: $$z_1\cdot z_2 =(a+bi)(a'+b'i)=(a\cdot a' -b\cdot b')+(a\cdot b'+b\cdot a')i$$$

For example: $$z_1=3+5i \$$, $$\ z_2=4+7i$$.

In this case we have $$a=3, \ b=5, \ a'=4, \ b'=7$$.

Then substituting in the formula: $$\displaystyle \begin{array}{rl} z_1\cdot z_2 &=(3+5i)(4+7i)=(3\cdot 4 -5\cdot 7)+(3\cdot 7+5\cdot 4)i \\ &=(12-35)+(21+20)i=-23+41i \end{array}$$$Let's see another example: $$z_1=9-15i \$$ and $$\ z_2=4+i$$. Now we have that $$a=9, \ b=-15, \ a'=4, \ b'=1$$. Then substituting in the formula we obtain: $$\displaystyle \begin{array}{rl} z_1\cdot z_2 &=(9-15i)(4+i)=(9\cdot 4 -(-15)\cdot 1)+(9\cdot 1+(-15)\cdot 4)i \\ &=(36+15)+(9-60)i=51-51i \end{array}$$$

Special cases:

What happens when we multiply pure imaginary numbers?

Let's remember that a pure imaginary number is a complex number of the form $$a+bi$$ where $$a=0$$ and, therefore, it is in fact a multiple of the imaginary unit $$i$$.

Let's see an example: $$5i\cdot9i=9\cdot5\cdot i^2= 45\cdot (-1)= -45$$$We obtain a real number, or a complex number with an imaginary part equal to zero. So the result is a complex number $$a+bi$$ where $$b=0$$. Therefore, whenever we have a product of two pure imaginary numbers, the result must be a real number. Other examples: $$\sqrt{-16}\cdot(-8i)=\sqrt{16}\cdot\sqrt{-1}(-8i)=4i\cdot(-8i)=-32i^2=32$$ $$\dfrac{3}{2}i\cdot 60i=90i^2=90\cdot(-1)=-90$$ As in the product of real numbers, a neutral element exists, which is $$1$$. The neutral element of the product is the complex $$z=1+0i$$, that is to say $$z=1$$. As we see, it coincides with the real number $$1$$, which is the neutral element of real numbers. Therefore, any complex number $$a+bi$$ multiplied by $$1+0i=1$$ will give us the same number $$a+bi$$. $$(4-6i)\cdot(1+0i)=(4-6i)\cdot 1=4-6i$$$

Let's see how to proceed to solve quotients of complex numbers expressed in binomic form. Let's have an example to better understand what to do. Let's look for the solution of the quotient:

$$\dfrac{4+3i}{2+i}$$$The first thing that we realize is that when we have $$i$$ in the denominator this is something what we cannot work with. That's why the first thing that we will do is to make this element disappear from the denominator. To do so, notice that if we multiply a complex number $$a+bi$$ by the complex $$a-bi$$, by means of the formula of the product we will obtain: $$(a+bi)\cdot(a-bi)=a^2⁻b^2\cdot i^2=a^2+b^2$$$

Therefore, we obtain a number without an imaginary part; in other words, multiplying by $$(a-bi)$$ we eliminate the imaginary component $$i$$.

That's why, comming back to the example, to eliminate $$i$$ from the denominator (where we have $$2 + i$$), what we do is to multiply the denominator by $$2-i$$. This is: $$\dfrac{4+3i}{2+i} \cdot \dfrac{1}{2-i}$$$But since we are in fact dividing by $$2-i$$, we will simultaneously multiply by $$2-i$$ not to change the result of the operation. Then: $$\dfrac{4+3i}{2+i}= \dfrac{4+3i}{2+i} \cdot \dfrac{2-i}{2-i}$$$

If we do each of the products we have:

$$(4+3i)\cdot(2-i)=(4\cdot2-3\cdot(-1))+(4\cdot(-1)+3\cdot2)i=$$

$$=(8+3)+(-4+6)i=11+2i$$

$$(2+i)\cdot(2-i)=2^2-i^2=4-(-1)=5$$

Therefore the quotient is now:

$$\dfrac{4+3i}{2+i}= \dfrac{4+3i}{2+i} \cdot \dfrac{2-i}{2-i}= \dfrac{11+2i}{5}$$$Now, since in the denominator there is no term with $$i$$, we can simply separate the quotient in two parts, $$\dfrac{11+2i}{5}=\dfrac{11}{5}+\dfrac{2}{5}i$$$ there is now a complex number with a real part $$\dfrac{11}{5}$$ and an imaginary part $$\dfrac{2}{5}$$.

Once we have seen this with an example, let's go to the general form.

To learn how to solve quotients of complex numbers, first we will define what the conjugate of a complex number is. Given a complex number $$z=a+bi$$ its conjugate is the number $$\bar{z}=a-bi$$. Namely we change the sign of the imaginary part. It is represented by means of a line over the complex number, $$\bar{z}$$.

If the imaginary part is positive, the conjugate is negative.

If $$z=8+5i$$ then the conjugate will be $$\bar{z}=8-5i$$

If the imaginary part is negative, the imaginary part of the conjugate is positive.

If $$z=8-5i$$ then the conjugate will be $$\bar{z}=8-(-5)i=8+5i$$

Once we have defined the conjugate of a complex number, we can generalize as to how to solve quotients of complex numbers.

Given two complex numbers $$z_1=a+bi$$$$$z_2=a'+b'i$$$

its quotient is calculated as follows:

First we must eliminate $$i$$ form the denominator. We multiply and divide the quotient by the conjugate of the denominator, so that it eliminates the imaginary part. That is: $$\dfrac{a+bi}{a'+b'i}=\dfrac{a+bi}{a'+b'i}\cdot \dfrac{a'-b'i}{a'-b'i}$$$Since the denominator is $$a'+b'i$$ its conjugate will be $$a'+b'i$$. Now, we do the products that we have, both in the numerator and in the denominator, by means of the procedure previously explained. This way we obtain: $$\dfrac{a+bi}{a'+b'i}=\dfrac{a+bi}{a'+b'i}\cdot \dfrac{a'-b'i}{a'-b'i} = \dfrac{(a\cdot a'-b\cdot b')+(a\cdot b'+b\cdot a')i}{(a')^2+(b')^2}$$$

Once we have done these operations, we just have to separate the fraction in two parts, the real one and the imaginary part. We already have the complex number resulting from the quotient of the two complex numbers of the example.

If $$z_1=3+5i$$ and $$z_2=4+7i$$

The denominator is $$4+7i$$, so the conjugate will be $$4-7i$$.

First we multiply and divide by this complex number: $$\dfrac{3+5i}{4+7i}=\dfrac{3+5i}{4+7i}\cdot \dfrac{4-7i}{4-7i}$$$We do the products in the numerator and the denominator: $$\dfrac{3+5i}{4+7i}=\dfrac{3+5i}{4+7i}\cdot \dfrac{4-7i}{4-7i} = \dfrac{(3\cdot4+5\cdot7)+(3\cdot(-7)+5\cdot4)i}{4^2+7^2}$$$

Joining and adding the terms we have: $$\dfrac{3+5i}{4+7i}=\dfrac{3+5i}{4+7i}\cdot \dfrac{4-7i}{4-7i} = \dfrac{(3\cdot4+5\cdot7)+(3\cdot(-7)+5\cdot4)i}{4^2+7^2}= \dfrac{47-i}{65}$$$If we separate the fraction in the real and the imaginary parts, we obtain the complex number resulting from the quotient in the binomic form: $$\dfrac{3+5i}{4+7i}=\dfrac{47}{65}-\dfrac{1}{65}i$$$

Let's remember that it is not possible to divide by the complex number $$0$$, because it does not make sense. It is completely prohibited. What we can do is to divide by a complex number that only has a real part, or only an imaginary part, but never with one that has both parts equal to zero.

For example: $$\dfrac{4+18i}{2}=\dfrac{4}{2} + \dfrac{18}{2}i = 2+9i$$$In this a case we divide by a complex number that only has real part, so it is a real number. Therefore its conjugate is itself. $$\dfrac{6-18i}{2i}=\dfrac{6-18i}{2i}\cdot \dfrac{-2i}{-2i} = \dfrac{6\cdot(-2i)+18\cdot 2i^2}{4}= \dfrac{-12i-36}{4}=-9-3i$$$ In this case we divide by a pure complex number. Its conjugate is itself but the sign changes.

To work with the quotient we have defined the conjugate of a complex number. But there also exists what is called the opposite of a complex number. In this case, what we change of the sign is not just the imaginary part (as in the conjugate), but also the real part. Therefore, if we have a complex number $$z=a+bi$$, its opposite number will be: $$-z=-(a+bi)=-a-bi$$.

For example: Opposite of $$6+3i$$ is $$-6-3i$$.

Opposite of $$6-3i$$ is $$-6+3i$$.

Opposite of $$-6-3i$$ is $$+6+3i$$.

Opposite of $$-6+3i$$ is $$6-3i$$.