Question

Square root of complex no. 16+30i

Answer 1

Answer:

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)

therefore,

X = ± 5 , Y =±3

=> ± (5–3i) • answer... ✌✌

★ mark as brainliest... ❤❤

Answer 2

Answer:

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.

: )