Question

How do you simplify – 4i. 5i?

Answer 1

Hint: We are given two terms – 4i and 5i and we are asked to produce these two terms. To do so we will understand what are complex numbers, what do i stand for and we then also learn what happens if i gets multiplied by i. We will use that \[{{i}^{2}}=-1,{{i}^{3}}=-1,{{i}^{4}}=1.\] Then we see that when we multiply the term constant, multiplied by each other and iota (i) will rise.

Complete step by step answer:
We are given two terms – 4i and 5i, these two contain i. We will learn what type of number contains i and what i means. Now there is a number type called the complex number which is represented as a + ib where a and b are real numbers and i stand for iota and i is denoted for \[\sqrt{-1}.\] Using this we can see that \[{{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1,{{i}^{3}}={{\left( \sqrt{-1} \right)}^{3}}=-i,{{i}^{4}}=1.\] Now in complex term a + ib, when we add we can add the real part with real and imaginary part with imaginary part.
\[z=\underbrace{a}_{\text{real part}}+\underbrace{ib}_{\text{imaginary part}}\]
When we multiply two complex numbers, we will multiply as usually we do as terms will be multiplied by another term. But in the later part, we need to simplify using the rule of addition. For example we have (2 + 3i) + (2 + 5i). The 2 will be added 2 and 3i will be added to 5i. So, (2 + 3i) (2 + 5i) will become 4 + 8i. When we multiply (2 + 3i) (2 + 5i), 2 will be multiplied by while 2 + 5i and 3i will be multiplied by while 2 + 5i. So, we get,
\[\left( 2+3i \right)\left( 2+5i \right)=2\left( 3+5i \right)+3i\left( 2+5i \right)\]
\[\Rightarrow \left( 2+3i \right)\left( 2+5i \right)=6+10i+6i+15{{i}^{2}}\]
As \[{{i}^{2}}=-1,\] so we get,
\[\Rightarrow \left( 2+3i \right)\left( 2+5i \right)=6+10i+6i-15\]
Using addition rule, we get,
\[\Rightarrow \left( 2+3i \right)\left( 2+5i \right)=-9+16i\]
Now in our problem we have \[-4i\times 5i.\] They just have imaginary parts. So, multiplication is much easier. So, we get,
\[-4i\times 5i=-4\times 5\times i\times i\]
\[\Rightarrow -4i\times 5i=-20{{i}^{2}}\]
As \[{{i}^{2}}=-1,\] so we get,
\[\Rightarrow -4i\times 5i=-20\times -1\]
\[\Rightarrow -4i\times 5i=20\left[ \text{As }-1\times -1=1 \right]\]

Note:
Remember to be careful around adding. One can add an imaginary part to the real part which makes things wrong. Also, if we have say the addition of 2 and 3i, so we just simplify and write 2 + 3i. No further can be solved. Also, remember that the product of 2 imaginary things is always real. So, we can cross-check.