Question

Find the square root of -25i.

Answer 1

A number.

Answer 2

It could be interpreted as an expression; you may have thought i was a variable.

Answer 3

For this,
Let:

\[-25i = x + iy\]

Now tell me what is x here and what is y here by comparing..

Answer 4

x is the integer, y is the complex part of of number to be multiplied by i. I have i = a^2 +2iab - b^2 and a^2 - b^2 = 0.

Answer 5

After that I am stuck.

Answer 6

What you get as 2ab??

Answer 7

Squaring both sides.

Answer 8

I am using a+bi. I suppose I should have posted it in x+yi form.

Answer 9

Let:

\[\sqrt{-25i} = a + ib\]
Squaring both the sides:

\[-25i = (a^2-b^2) + 2abi\]

From here:

\((a^2 - b^2) = 0\) -----------1

\[2ab = -25\]

Square it:

\[4a^2.b^2 = 625\] -----------2

Got till here??

Answer 10

I see. So a = 25/2

Answer 11

No a is not that much..

Use this formula:

\[(a^2 + b^2)^2 = (a^2-b^2)^2 + 4a^2b^2\]

Can you find \(a^2 + b^2\) from here??

Answer 12

Where did that come from?

Answer 13

Oh wait, is a = 5/sqrt(2) ?

Answer 14

Yes you find a correct..
Now find b..

Answer 15

I see.

Answer 16

\[(a^2 - b^2)^2 = a^4 + b^4 - 2a^2.b^2\]

Now add \(4a^2.b^2\) both the sides,

\[(a^2-b^2)^2 + 4a^2.b^2 = a^2 + b^4 + 2a^2.b^2 \implies (a^2+b^2)^2\]

This is what I have written there..

Answer 17

I found b to be (5i)/sqrt(2).

Answer 18

This one is wrong..
Show me how you did it??

Answer 19

I mean b = 5/sqrt(2). Of a = 5/sqrt(2), then that minus itself gives 0.

Answer 20

Yes now you are right..
Both the values of a and b will be same..
But there is one little mistake:

\[a = \pm \frac{5}{\sqrt{2}} \qquad Or \qquad b = \pm \frac{5}{\sqrt{2}}\]

Answer 21

I see. Thank you. That setting the complex number to a+bi is a clever thing.

Answer 22

Um, @waterineyes you know it simplifies easier than that, right? I'm using a different technique than what you're doing, mind posting what you got?

Answer 23

In simplified form I mean, I want to see if it matches what I got

Answer 24

(-1)^something style

Answer 25

After that you will do like this:

\[\sqrt{-25i} = \pm (\frac{5}{\sqrt{2}} + i \frac{5}{\sqrt{2}})\]

Answer 26

\[\sqrt{-25i} = \sqrt{-1*5*5*\sqrt{-1}} = 5 * \sqrt{-1 \sqrt{-1}}=5(-1(-1)^{1/2})^{1/2}=-5 (-1)^{3/4}\]

Answer 27

Is this the square root of this??

Answer 28

http://www.wolframalpha.com/input/?i=sqrt%28-25i%29+%3D+-5+%28-1%29%5E%283%2F4%29

Answer 29

You have done correctly there is no doubt that this is wrong but I have learned to take the square root of complex number as I did..
Sorry..

Answer 30

Yep, I'm correct.
http://www.wolframalpha.com/input/?i=square+root+of+%28-25i%29

Answer 31

Wolfram is not everything..
There are certain algorithms are made to solve particular things and you must follow that..

Answer 32

I just used fractional exponents and treated -1 as if it was a variable for a moment, that's how I did that in my head.

Answer 33

Think about it, what's \(\large\sqrt{\sqrt{x}}\) ?
x\(^{1/4}\), agreed?

What's 1-(1/4)? 3/4.
Where did that 1 go? Well, kind of like this: (\(\sqrt{-1}\))^2 = \(((-1)^{2/4})^{4/2}\) = -1

I am following what you've said waterineyes, it's just unusual to me because I did it with a completely different method.

Answer 34

That's why I asked what you got with your method :-) There's often more than one correct way of solving something in mathematics after all