Question

Express the given complex number in the form \[a+ib\]: \[\left( 5i \right)(-\dfrac{3}{5}i)\].

Answer 1

Hint: You need to know the rules for the algebra of complex numbers and the behaviour of iota\[\left( i \right)\] when raised to a certain power in order to solve this question.

Complete step by step solution:
A complex number is a number with two parts: Real and imaginary parts. General form of the complex number is:
\[a+ib\]
Where a is the real part and b is the imaginary part.
Iota \[\left( i \right)\] is defined as the square root of \[-1\].
\[i=\sqrt{-1}\]
 A complex number has two parts as mentioned above and either of them can be zero. So all real numbers are complex numbers and all imaginary numbers are also complex numbers.
Let’s see how does the iota \[\left( i \right)\] behave when raised to power,
\[i=\sqrt{-1}\]
\[\begin{align}
  & {{i}^{0}}=1 \\
 & {{i}^{1}}=i \\
 & {{i}^{2}}=-1 \\
 & {{i}^{3}}=-i \\
 & {{i}^{4}}=1 \\
\end{align}\]
Clearly, we can see the behaviour of \[i\] and the pattern repeats itself in a cycle of four.
Therefore, we can generalise it to
\[\begin{align}
  & {{i}^{4k}}=1 \\
 & {{i}^{4k+1}}=i \\
 & {{i}^{4k+2}}=-1 \\
 & {{i}^{4k+3}}=-i \\
\end{align}\]
Now, let’s come to our question,
Given: \[\left( 5i \right)(-\dfrac{3}{5}i)\]
The multiplication is done in the same way as done in the real numbers. Multiply the real number with the real number and imaginary number with the imaginary number.
\[5\times \dfrac{-3}{5}\times i\times i=-3{{i}^{2}}=3=3+0i\] (\[\because \]\[{{i}^{2}}=-1\] )
We can compare it with the general form of complex numbers. It gives that a\[=\]\[3\]and b\[=0\].
Therefore, \[\left( 5i \right)(-\dfrac{3}{5}i)\] $=3+0i$. It means that the imaginary part of this complex number is zero and it is a purely real number equal to \[3\].

Note:
If the real part of a complex number is zero, it is called a purely imaginary number. For example, \[2i,-9i,\dfrac{5}{3}i\] and if the imaginary part of a complex number is zero, it is called a purely real number as seen in the above solution.