Question

If ${z_1} = \sqrt 2 \left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)$ and ${z_2} = \sqrt 3 \left( {\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}} \right)$ , then $\left| {{z_1}{z_2}} \right|$ is equal to
A. $6$
B. $\sqrt 2 $
C. $\sqrt 6 $
D. $\sqrt 2 + \sqrt 3 $

Answer 1

Hint: In this question we have been given ${z_1} = \sqrt 2 \left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)$ and ${z_2} = \sqrt 3 \left( {\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}} \right)$We have the value of $\left| {{z_1}{z_2}} \right|$.
we can see that the value is in mode and the above expression is in the form of
$z = x + iy$ . This is the form of a complex number. So if we have to find the value of
$\left| z \right|$ , we can write the expression as the root of the square of the real number which is $x$ and the root of the square of the imaginary number which is $y$, i.e.
$\left| z \right| = \sqrt {{x^2} + {y^2}} $ . So we will use this formula to solve the above question.

Complete step by step solution:
Here we have
${z_1} = \sqrt 2 \left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)$
And,
 ${z_2} = \sqrt 3 \left( {\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}} \right)$ ,
In the first term, we have real number i.e.
$x = \cos \dfrac{\pi }{4}$ and the imaginary number i.e.
$y = \sin \dfrac{\pi }{4}$
We can also write the term as
${z_1} = \left( {\sqrt 2 \cos \dfrac{\pi }{4} + \sqrt 2 i\sin \dfrac{\pi }{4}} \right)$
So by applying the formula we can write
 $\left| {{z_1}} \right| = \sqrt {\left( {\sqrt 2 \cos \dfrac{\pi }{4}}^2 \right) + {{\left( {\sqrt 2 \sin \dfrac{\pi }{4}} \right)}^2}} $
We can take the common factor out, so we have
 $\left| {{z_1}} \right| = \sqrt {{{(\sqrt 2 )}^2}{{\left( {\cos \dfrac{\pi }{4}} \right)}^2} + {{\left( {\sin \dfrac{\pi }{4}} \right)}^2}} $
We know that
${\sin ^2}\theta + {\cos ^2}\theta = 1$ , so by applying this we can write
 $\left| {{z_1}} \right| = \sqrt 2 \times 1 = \sqrt 2 $
Now we will solve
 ${z_2} = \sqrt 3 \left( {\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}} \right)$ ,
Here we have real number i.e.
$x = \cos \dfrac{\pi }{3}$ and the imaginary number i.e.
$y = \sin \dfrac{\pi }{3}$
We can also write the term as
${z_2} = \left( {\sqrt 3 \cos \dfrac{\pi }{3} + \sqrt {3\,} i\sin \dfrac{\pi }{3}} \right)$
Similarly as above applying the formula we can write
$\left| {{z_2}} \right| = \sqrt {\left( {\sqrt 3 \cos \dfrac{\pi }{3}}^2 \right) + {{\left( {\sqrt 3 \sin \dfrac{\pi }{3}} \right)}^2}} $
We can take the common factor out, so we have
 $\left| {{z_2}} \right| = \sqrt {{{(\sqrt 3 )}^2}{{\left( {\cos \dfrac{\pi }{3}} \right)}^2} + {{\left( {\sin \dfrac{\pi }{3}} \right)}^2}} $
Now by applying the identity of
${\sin ^2}\theta + {\cos ^2}\theta = 1$ , we can write
 $\left| {{z_2}} \right| = \sqrt 3 \times 1 = \sqrt 3 $
Now we can put the values together and we have
 $\left| {{z_1}{z_2}} \right| = \sqrt 2 \times \sqrt 3 = \sqrt 6 $
Hence the correct option is (C) $\sqrt 6 $.

Note:
We should note that the value of $i$ is $\sqrt { - 1} $ .
This is the imaginary number, also called iota. So if we square the value of iota i.e. ${i^2}$ , we get the value $\sqrt { - 1} \times \sqrt { - 1} = - 1$ .
So it gives us ${i^2} = - 1$ .