Question

If z = 1 + i, then find
${{z}^{2}}$

Answer 1

- Hint: Substitute the value of z into the equation and then solve it like a normal algebraic equation to find the square of z.
You can use distributive property:
b.(a + c) = b.a + b.c

Complete step-by-step solution -

First I will say the proper definition for a complex number
A complex number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i is a solution of the equation:
${{x}^{2}}=-1.....\left( 1 \right)$
Because no real number satisfies this equation, i is called an imaginary number.
Now we need the value of the square of z.
We can write square of z as:
${{z}^{2}}=z\times z$
So by substituting the value of z = 1 + i, we get:
(1+i).(1+i)
By assuming (1 + i) as one whole entity and applying distributive law, we get:
By distributive property:
b.(a + c) = b.a + b.c
(1 + i).(1 + i) = 1.(1 + i) + i.(1 + i)
Now again applying the distributive law two times, we get:
= 1 + i + i + i.i
By equation (1), we can say:
$i.i={{i}^{2}}=-1$
By substituting this, we get:
(1 + i).(1 + i) = 1 + i + i -1
By cancelling common terms, we get:
(1 + i).(1 + i) = i + i
By simplifying, we get:
(1 + i).(1 + i) = 2i
By substituting 1 + i as z, we get:

& z.z=2i \\\ & {{z}^{2}}=2i \\\ \end{aligned}$$ _$$\therefore $$The required value is 2i_ Note: Alternative method- This method is easy and quicker. The process is to multiply directly the terms by using algebraic identity: $${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$$ After applying this identity, use equation (1) which is given by i.i = -1. So substitute the value -1 wherever you see the term i.i. After substituting the value -1, the leftover equation is normal calculation, so do it algebraically to reach a solution.