Question

How to find $\operatorname{Re} :z$, when $z = {i^{i + 1}}$?

Answer 1

In this question, we need to find the real part of the given complex number. Note that a complex number is a combination of a real number and an imaginary number. Firstly we will use the multiplication rule of exponents which is given by ${a^m} \cdot {a^n} = {a^{m + n}}$. We find the value of $i$ in terms of ${e^{ix}}$ for a suitable value of x. We take here $x = \dfrac{\pi }{2}$. Then we simplify further to obtain the real part of the given complex number.

Complete step-by-step answer:
Given a complex number of the form $z = {i^{i + 1}}$
We are asked to determine the real part of this complex number.
A complex number is represented as $z = a + ib$, where a and b are real numbers and $i$ is an imaginary unit whose value is $i = \sqrt { - 1}$.
Now consider the complex number $z = {i^{i + 1}}$ …… (1)
We have the multiplication rule of exponents which is given by,
${a^m} \cdot {a^n} = {a^{m + n}}$
Here $m = i$ and $n = 1$
Hence we have,
$\Rightarrow {i^{i + 1}} = {i^i} \cdot {i^1}$
This can also be written as,
$\Rightarrow {i^{i + 1}} = {i^i} \cdot i$ …… (2)
We have the Euler’s identity given by, ${e^{ix}} = \cos x + i\sin x$
For $x = \dfrac{\pi }{2}$, we have the identity as,
$\Rightarrow {e^{i\dfrac{\pi }{2}}} = \cos \dfrac{\pi }{2} + i\sin \dfrac{\pi }{2}$
We know that $\cos \dfrac{\pi }{2} = 0$ and $\sin \dfrac{\pi }{2} = 1$.
Hence substituting this we get,
$\Rightarrow {e^{i\dfrac{\pi }{2}}} = 0 + i(1)$
$\Rightarrow {e^{i\dfrac{\pi }{2}}} = i$
Now substituting $i = {e^{i\dfrac{\pi }{2}}}$ in the equation (2) we get,
$\Rightarrow {i^{i + 1}} = {\left( {{e^{i\dfrac{\pi }{2}}}} \right)^i} \cdot i$
This can also be written as,
$\Rightarrow {i^{i + 1}} = {e^{\left( {i\dfrac{\pi }{2}} \right)i}} \cdot i$
$\Rightarrow {i^{i + 1}} = {e^{\left( {\dfrac{\pi }{2}} \right){i^2}}} \cdot i$
We know the value of $i$ which is given by $i = \sqrt { - 1}$, then ${i^2} = - 1$.
Hence we get,
$\Rightarrow {i^{i + 1}} = {e^{ - \dfrac{\pi }{2}}} \cdot i$, which is purely imaginary.
Hence there is no real part. So the real part will be equal to zero.
Hence for the complex number $z = {i^{i + 1}}$, we have $\operatorname{Re} (z) = \operatorname{Re} ({i^{i + 1}}) = 0$.

Note:
A complex number is a combination of a real number and an imaginary number. Any number in mathematics can be known as a real number. Imaginary numbers are the numbers which when squared give a negative number.
A complex number is represented as $z = a + ib$, where a and b are real numbers and $i$ is an imaginary unit whose value is $i = \sqrt { - 1}$.
Students must remember the multiplication rule of exponents which is given by,
${a^m} \cdot {a^n} = {a^{m + n}}$
Also the Euler’s identity given by ${e^{ix}} = \cos x + i\sin x$
By using this identity we have, ${e^{i\dfrac{\pi }{2}}} = i$, ${e^{i\pi }} = - 1$ and ${e^{i2\pi }} = 1$.