Question

Find the Square root of
i)1-2i

Answer 1

Answer:

Solve the equation

(a+ib)2=(a2−b2)+i(2ab)=1+2i.

where a,b∈R. Hence b=1/a and 1=a2−b2=a2−1/a2, that is

a4−a2−1=0.

Can you take it from here?

At the end you should find that one root is z1=(5–√+1)/2−−−−−−−−−√+i(5–√−1)/2−−−−−−−−−√ and the other one is z2=−z1.

Answer 2

The Square root of 1-2i   is   $\±( \sqrt{\frac{\sqrt{5} +1}{2} } +\sqrt{\frac{\sqrt{5} -1}{2} } \ i)$.

Step-by-step explanation:

Given:

1-2i

To Find:

The Square root of 1-2i.

Formula Used:

$(x+y)^{2} =x^{2} +y^{2} +2xy$

$(x^{2} +y^{2} )^{2} =(x^{2} -y^{2} )^{2} +4x^{2} y^{2}$

$i=\sqrt{-1}$

Solution:

As given,1-2i.

Let $\sqrt{ 1-2i} =a+ib$

Squaring both sides.

$(\sqrt{ 1-2i})^{2} =(a+ib)^{2} \\ 1-2i = a^{2} +(ib)^{2} +i(2ab)\\ 1-2i= (a^{2} -b^{2} )+i(2ab)$  ------------- equation no.01.

Comparing real and imaginary parts  of equation no.01,we get.

$a^{2} -b^{2} =1$   ------------ equation no.02.

$\\ \ 2ab=-2$

$(a^{2} +b^{2} )^{2} =(a^{2} -b^{2} )^{2} +4a^{2} b^{2}$

               $=1^{2} +(-2)^{2} \\ =1+4 =5$

$a^{2} +b^{2} =\sqrt{5}$   ---- equation no.03.

Adding  equation no.02 and equation no.03 ,we get.

$2a^{2} =\sqrt{5}+1$

$a= \± \sqrt{\frac{\sqrt{5} +1}{2} }$

Subtracting equation no.03 from equation no.02,we get.

$-2b^{2} =1-\sqrt{5}$

$2b^{2} =\sqrt{5}-1$  

$b = \± \sqrt{\frac{\sqrt{5} -1}{2} }$

Therefore,

$\sqrt{ 1-2i} =a+ib= \±( \sqrt{\frac{\sqrt{5} +1}{2} } + \sqrt{\frac{\sqrt{5} -1}{2} } \ i)$

Thus,the Square root of 1-2i   is   $\±( \sqrt{\frac{\sqrt{5} +1}{2} } + \sqrt{\frac{\sqrt{5} -1}{2} } \ i)$.

#SPJ3