Question

The square root of -8i is?

Answer 1

8^.5(1+i)

Answer 2

sorry it's wrong

Answer 3

yeah 2(1+i)

Answer 4

The answer is \[\pm 2(1-i)\] please explain ,how?

Answer 5

1-i?..r u sure?

Answer 6

yes i am damn sure

Answer 7

pellet..its -8i...sorry..i did for 8i..yeah its correct

Answer 8

How> explain the steps?

Answer 9

I'll telll u the easiest process...take a+ib to be the root..square it then equate the real and imaginary parts with -8i...find a and b

Answer 10

I tried the same method....but i didn't get it. please write the steps you have taken.

Answer 11

(a+ib)^2=-8i
a^2+2iab-b^2=-8i..
now real part is 0..so a^2=b^2...meaning a=+/- b
equating imaginary parts
2ab=-8
now if a=b, then 2a^2=-8..thats not possible( and b are obviously real)
so a=-b
so -2a^2=-8
a=2
b=-2
so 2(1-i)

Answer 12

or else try using eulers method..and de moivre's theorem...thats the fastest process

Answer 13

Thanks @amriju

Answer 14

w.c. :)

Answer 15

:)

Answer 16

Euler is much easy.....

Answer 17

yup..and its fastest...

Answer 18

ok one more please....

Answer 19

okk..n.p

Answer 20

i'll try

Answer 21

square root of 3+4i?? before one i tried from euler formula i got the answer but this won't come from that.

Answer 22

why don't you try a+ib..? ok..if u want to try euler
5(3/5+4/5i) u know how I did it..dont u?
5 (cos 53 +i sin 53)
5e^(i*53pi/180)
sq. root it
sq.root(5)*e^(i*53pi/360)
now convert it into cos x + isin x form..and solve

Answer 23

ok i got it...thanks
ok can you give me procedure of a+ib .... i am confused somewhat in it.

Answer 24

the first step is to assume a root of the form a+ib
square it to get a^2+2iab -b^2
now look at the rhs...it has a real and a complex part, rite
real and complex parts never mess with one another..so a^2-b^2=real part in the rhs
2iab=complex part of rhs..
two variables, two eqns..solve it for a and b

Answer 25

can you do it now?

Answer 26

If i have some, i will ask... thanks btw