Math, asked by nmohd7700, 1 year ago

Find the square root of 16-30i

Answers

Answered by mysticd
111
Hi ,

Let √( 16 - 30i ) = x - iy ---( 1 )

Do the square of equation ( 1 ) , we

get ,

16 - 30i = ( x - iy )²

= x² + ( iy )² - 2 × x × iy

= ( x² - y² ) - 2xyi

Compare both sides ,

x² - y² = 16 ---( 2 )

2xy = 30

y = 30/2x

y = 15/x ---( 3 )

Substitute y value in equation ( 2 ),

we get

x² - ( 15/x )² = 16

x² - 225/x² = 16

x⁴ - 225 = 16x²

x⁴ - 16x² - 225 = 0

x⁴ - 25x² + 9x² - 225 = 0

x² ( x² - 25 ) + 9 ( x² - 25 ) = 0

( x² - 25 ) ( x² + 9 ) = 0

x² - 25 = 0 or x² + 9 = 0

x² = 25

x = ± 5

Substitute x = ± in equation ( 3 ) ,

We get ,

y = 15/( ± 5 )

y = ± 3

Therefore ,

x = ± 5 , y = ± 3

√( 16 - 30 i ) = x - iy

= ± ( 5 - 3i )

I hope this helps you.

: )



Answered by Anonymous
16

\textbf{\underline{\underline{According\:to\:the\:Question}}}

Assumption

√16 - 30i = (x + iy) ..... (1)

\textbf{\underline{Squarring\;both\;sides :-}}

(16 - 30i) = (x - iy)²

(16 - 30i) = (x² - y²) - i(2xy) ... (2)

Here

\Large{\boxed{\sf\:{Real\;and\;imaginary\;part\;of\;both\;sides\;of\;(2)}}}

(x² - y²) = 16

2xy = 30

(x² + y²) = √{(x² - y²)² + 4x²y²}

= √(16² + 30²)

= √(256 + 900)

= √1156

= 34

Hence,

(x² - y²) = 16 .... (3)

Also,

(x² + y²) = 34 ..... (4)

\Large{\boxed{\sf\:{Solving\;(3)\;and\;(4)}}}

x² = 25

x = √25

x = ±5

Also,

y² = 9

y = √9

y = ±3

Here

xy > 0

Therefore,

(x = 5 and y = 3) or (x = -5 and y = -3)

Hence,

√16 - 30i = (5 - 3i) or (-5 + 3i)

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