Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

How do you simplify ${{\left( 2+5\iota \right)}^{2}}$?

seo-qna
SearchIcon
Answer
VerifiedVerified
424.2k+ views
Hint: In this problem, we have to find the square of a complex number. This can also be done by multiplying the complex number by itself. After multiplication, we shall group all the terms and further simplify them. Then we will use the basic properties of complex numbers such as the different values of iota when it is raised to various powers.

Complete step-by-step answer:
In order to simplify the given expression, we must have prior knowledge of complex numbers. A complex number is of the form, $x+\iota y$. It comprises two parts. One is the real number part which lies on the x-axis of the cartesian plane and the other is a complex number part which lies on the y-axis of the cartesian plane.
The expression given can also be written as:
${{\left( 2+5\iota \right)}^{2}}=\left( 2+5\iota \right)\left( 2+5\iota \right)$
Thus, we shall multiply the complex function by itself.
$\Rightarrow {{\left( 2+5\iota \right)}^{2}}=4+10\iota +10\iota +25{{\iota }^{2}}$
Adding the like terms of iota, we get
$\Rightarrow {{\left( 2+5\iota \right)}^{2}}=4+20\iota +25{{\iota }^{2}}$
There are predefined values assigned to iota when it is raised to certain powers. They are as follows.
${{\iota }^{1}}=\iota $
${{\iota }^{2}}=-1$
${{\iota }^{3}}=-\iota $
${{\iota }^{4}}=1$
From the above results, we see that ${{\iota }^{2}}=-1$. Substituting this value in our equation, we get
$\Rightarrow {{\left( 2+5\iota \right)}^{2}}=4+20\iota +25\left( -1 \right)$
 $\begin{align}
  & \Rightarrow {{\left( 2+5\iota \right)}^{2}}=4+20\iota -25 \\
 & \Rightarrow {{\left( 2+5\iota \right)}^{2}}=-21+20\iota \\
\end{align}$
Therefore, the given expression ${{\left( 2+5\iota \right)}^{2}}$ is simplified to $-21+20\iota $.

Note:
The generalized rule for iota raised to any power is that iota raised to the power of 4 or multiples of 4 is equal to 1. Otherwise, the value of iota in terms of multiples 4 is given as, ${{\iota }^{4k+1}}=\iota $, ${{\iota }^{4k+2}}=-1$ and ${{\iota }^{4k+3}}=-\iota $ where $k$ is a constant. Using this, we can easily find the value of iota even when its power is very large natural numbers.