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Question: The value of \(\sqrt {2i} \): A) \(1 + i\) B) \( - 1 - i\) C) \( - \sqrt {2i} \) D) None of ...

The value of 2i\sqrt {2i} :
A) 1+i1 + i
B) −1−i - 1 - i
C) −2i- \sqrt {2i}
D) None of these

Explanation

Solution

As the root of the complex number will also be a complex number, let the root of 2i2i as x+iyx + iy. Then, the equation will be 2i=x+iy\sqrt {2i} = x + iy. Square on both sides and compare the real and imaginary part of the equation and form equations. Solve the equations to find the value of xx and yy. Then, substitute the values in 2i=x+iy\sqrt {2i} = x + iy to find the required value.

Complete step by step solution: Since the given number is a complex number, therefore the square root of this number will also be a complex number.
So, let the square root of the given number 2i2i as x+iyx + iy . Thus we can write
(x+iy)2=2i{\left( {x + iy} \right)^2} = 2i
On simplifying the above equation we get,
x2−y2+2ixy=2i (1){x^2} - {y^2} + 2ixy = 2i{\text{ }}\left( {\text{1}} \right)
The part of the complex number without the ii as coefficient is called the real part of the complex number, and the part of the complex number with ii as its coefficient is called the imaginary part of the complex number.
On comparing the real and imaginary part of the equation (1)\left( 1 \right), we get
x2−y2=0 \-2xy=2  {x^2} - {y^2} = 0 \\\ \- 2xy = 2 \\\
On simplifying the equation we get
x2−y2=0 (2) xy=1 (3)  {x^2} - {y^2} = 0{\text{ }}\left( {\text{2}} \right) \\\ xy = 1{\text{ }}\left( {\text{3}} \right) \\\
Find the value of x2+y2{x^2} + {y^2}using the formula (m2+n2)2=(m2−n2)2+4m2n2{\left( {{m^2} + {n^2}} \right)^2} = {\left( {{m^2} - {n^2}} \right)^2} + 4{m^2}{n^2}
(x2+y2)2=(x2−y2)2+(2xy)2{\left( {{x^2} + {y^2}} \right)^2} = {\left( {{x^2} - {y^2}} \right)^2} + {\left( {2xy} \right)^2}
Substituting the values from equation (2) and (3)\left( 2 \right){\text{ and }}\left( 3 \right) we get
(x2+y2)2=(0)2+(2(−1))2 =0+4 =4 x2+y2=2 (4)  {\left( {{x^2} + {y^2}} \right)^2} = {\left( 0 \right)^2} + {\left( {2\left( { - 1} \right)} \right)^2} \\\ = 0 + 4 \\\ = 4 \\\ {x^2} + {y^2} = 2{\text{ }}\left( {\text{4}} \right) \\\
Adding equation 22 and 44we get
x2−y2+x2+y2=0+2 2x2=2 x2=1 x=±1  {x^2} - {y^2} + {x^2} + {y^2} = 0 + 2 \\\ {\text{2}}{x^2} = 2 \\\ {x^2} = 1 \\\ x = \pm 1 \\\
Substituting the ±1 \pm 1 for xxin the equation 11.
xy=−1xy = - 1
When x=1x = 1
Then,
(1)y=1 ⇒y=1  \left( 1 \right)y = 1 \\\ \Rightarrow y = 1 \\\
Similarly, When x=−1x = - 1
Then,
(−1)y=1 ⇒y=−1  \left( { - 1} \right)y = 1 \\\ \Rightarrow y = - 1 \\\

Substituting the values of x,yx,y in the required complex number
x+iy=1+ix + iy = 1 + i and x+iy=−1−ix + iy = - 1 - i

Hence, option A and B are correct.

Note: We can also do this question by adding and subtracting 1 to 2i2i and we will get 2i=1+2i−12i = 1 + 2i - 1. The term can be rewritten as 12+2(1)i−12⇒12+2(1)i+i2{1^2} + 2\left( 1 \right)i - {1^2} \Rightarrow {1^2} + 2\left( 1 \right)i + {i^2}. Hence, 2i=(1+i)22i = {\left( {1 + i} \right)^2}. On taking square roots both sides, we will get, 2i=±(1+i)\sqrt {2i} = \pm \left( {1 + i} \right). Also, the square root of the complex number will also be a complex number.