How do you solve ${(1 + 4i)^2}$ ?

Hint:To solve the questions like this, the first step we need to do is try to solve those using identities. Here, the question is clearly in the form of ${(a + b)^2}$ if we consider a=1 and b=4i. Therefore, we will expand the identity which is ${(a + b)^2} = {a^2} + {b^2} + 2ab$. Putting the respective values of $a$ and $b$ we will get that the questions simplifies into $1 + 16{i^2} + 8i$. Now, we will put the value of ${i^2}$ which is equal to -1. Doing this will give us $1 + 16 \times - 1 + 8i$. Solving this form using BODMAS we will get the answer as $ - 15 + 2i$.

Complete step by step answer:
Here, the complex number we have is ${(1 + 4i)^2}$. Let’s assume, a=1 and b=4i. Therefore, we can write the question as:-
${(1 + 4i)^2} = {(a + b)^2}$
At this point, we will use the identity ${(a + b)^2}$which is given by
${(a + b)^2} = {a^2} + {b^2} + 2ab$
Now, using the above identity in our question we will get
${(a + b)^2} = {a^2} + {b^2} + 2ab \\
\Rightarrow{(1 + 4i)^2} = {1^2} + {(4i)^2} + 2 \times 1 \times 4i \\
\Rightarrow{(1 + 4i)^2}= 1 + 16{i^2} + 8i $
Now, we know that ${i^2} = - 1$
Putting the value of ${i^2}$ in our answer, we will get
$1 + 16{i^2} + 8i = 1 + {(16 \times \sqrt { - 1} )^2} + 8i \\
\Rightarrow 1 + ( - 16) + 8i \\
\Rightarrow 1 - 16 + 2i \\
\therefore - 15 + 2i $

Note:Always remember that the value of i .i.e., iota is $\sqrt { - 1} $ and not$ - \sqrt 1 $.If you mistakenly use$ - \sqrt 1 $ then the answer for many operations will come wrong.
Suppose you need to find the value of $ - i - i$, so when you put $i = - \sqrt 1 $. Your answer will be $i - i = - \sqrt 1 - \left( { - \sqrt 1 } \right) = - \sqrt 1 + \sqrt 1 = 0$ which is not the correct answer. Therefore, always remember to use the correct value for iota. Also, don’t put the value of $i$ in the equation since it won’t make much of a difference as we won’t be able to operate between a complex number and real number. Therefore, only use the value of ${i^2}$.