Question

find the square root of the complex number 1-i

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Square root of complex number = 1 - i

Let $\textsf{\sqrt{1 - i}}$ = x + iy

$\textsf{1 - i = {( x + iy)} ^{2} = {x} ^{2} - {y} ^{2} + 2xiy}$

Now, Comparing real and imaginary parts,

$\textsf{{x} ^{2} - {y} ^2}$ = 1 --> ( i )

2xy = - 1 --> ( ii )

$\textsf{{x} ^2 + {y}^{2}}$ = $\textsf{\sqrt({x} ^2 - {y} ^{2}) + {(2xy)}^{2}}$

$\implies{\textsf{\sqrt{{1}^{2} + {(-1)}^{2}}}}$

$\implies{\textsf{\sqrt{2}}}$ --> ( iii )

Adding equation ( i ) and ( iii ),

$\textsf{{x} ^{2} - {y} ^2}$ + $\textsf{{x} ^2 + {y}^{2}}$ = $\textsf{1 + \sqrt{2}}$

$\textsf{{(2x)}^{2}}$ = $\textsf{1 + \sqrt{2}}$

$\textsf{{x}^{2}}$ = $\textsf{\dfrac{1 + \sqrt{2}}{2}}$

Putting value of $\textsf{{x}^{2}}$ in equation ( i ),

$\textsf{\dfrac{1 + \sqrt{2}}{2}}$ - $\textsf{{y}^{2}}$ = 1

$\textsf{{y}^{2}}$ = $\textsf{\dfrac{1 + \sqrt{2}}{2}}-1$

➡️ Roots are

x = $\textsf{\sqrt{\dfrac{1 + \sqrt{2}}{2}}}$

y = $\textsf{\sqrt{\dfrac{ \sqrt{2}-1}{2}}}$