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Question: If \({z_1} = \sqrt 2 \left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)\) and \({z_2} = \...

If z1=2(cos⁡π4+isin⁡π4){z_1} = \sqrt 2 \left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right) and z2=3(cos⁡π3+isin⁡π3){z_2} = \sqrt 3 \left( {\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}} \right) , then ∣z1z2∣\left| {{z_1}{z_2}} \right| is equal to
A. 66
B. 2\sqrt 2
C. 6\sqrt 6
D. 2+3\sqrt 2 + \sqrt 3

Explanation

Solution

In this question we have been given z1=2(cos⁡π4+isin⁡π4){z_1} = \sqrt 2 \left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right) and z2=3(cos⁡π3+isin⁡π3){z_2} = \sqrt 3 \left( {\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}} \right)We have the value of ∣z1z2∣\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+iyz = x + iy . This is the form of a complex number. So if we have to find the value of
∣z∣\left| z \right| , we can write the expression as the root of the square of the real number which is xx and the root of the square of the imaginary number which is yy, i.e.
∣z∣=x2+y2\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
z1=2(cos⁡π4+isin⁡π4){z_1} = \sqrt 2 \left( {\cos \dfrac{\pi }{4} + i\sin \dfrac{\pi }{4}} \right)
And,
z2=3(cos⁡π3+isin⁡π3){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⁡π4x = \cos \dfrac{\pi }{4} and the imaginary number i.e.
y=sin⁡π4y = \sin \dfrac{\pi }{4}
We can also write the term as
z1=(2cos⁡π4+2isin⁡π4){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
∣z1∣=(2cos⁡π42)+(2sin⁡π4)2\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
∣z1∣=(2)2(cos⁡π4)2+(sin⁡π4)2\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⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 , so by applying this we can write
∣z1∣=2×1=2\left| {{z_1}} \right| = \sqrt 2 \times 1 = \sqrt 2
Now we will solve
z2=3(cos⁡π3+isin⁡π3){z_2} = \sqrt 3 \left( {\cos \dfrac{\pi }{3} + i\sin \dfrac{\pi }{3}} \right) ,
Here we have real number i.e.
x=cos⁡π3x = \cos \dfrac{\pi }{3} and the imaginary number i.e.
y=sin⁡π3y = \sin \dfrac{\pi }{3}
We can also write the term as
z2=(3cos⁡π3+3 isin⁡π3){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
∣z2∣=(3cos⁡π32)+(3sin⁡π3)2\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
∣z2∣=(3)2(cos⁡π3)2+(sin⁡π3)2\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⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 , we can write
∣z2∣=3×1=3\left| {{z_2}} \right| = \sqrt 3 \times 1 = \sqrt 3
Now we can put the values together and we have
∣z1z2∣=2×3=6\left| {{z_1}{z_2}} \right| = \sqrt 2 \times \sqrt 3 = \sqrt 6
Hence the correct option is (C) 6\sqrt 6 .

Note:
We should note that the value of ii is −1\sqrt { - 1} .
This is the imaginary number, also called iota. So if we square the value of iota i.e. i2{i^2} , we get the value −1×−1=−1\sqrt { - 1} \times \sqrt { - 1} = - 1 .
So it gives us i2=−1{i^2} = - 1 .