# Terry Lee Square Root (1 Viewer)

#### D3spair

##### New Member
Could anyone explain to me the 'Terry Lee' method used as shortcut to find square roots of a complex number?

#### vinlatte

##### Member
Okay, i understand it quite well now. It makes alot of sense. I'll do your example, -3+4i :

I don't know about you but you should be able to write down the two equations straight off, using -3+4i = a2-b2+2abi and equating the Re and Im parts to obtain:

a2-b2=-3
2ab=4

Now the first thing we do is check that the b in the complex number is even because this will correspond to the 2ab when we take out the two from the even number. So in this example b=4, so we write 2x2, which corresponds to 2ab and then if you cancel the two from both you get ab=2, so we find two numbers that multiply to give 2 (a and b), as well as the difference between their squares equals -3. Now we mentally work out that 12-22=-3 and the order here is important because if it was the other around we would get 3. equating the numbers in 12-22=-3 and a2-b2=-3, it should be obvious that a=1, and b=2. Therefore your answer is a+bi => 1+2i.
But it is a square root so we need to take the positive and negative of it, so the final answer is +(1+2i).
This can actually be all done in your head, so some working might allow us to use this in an exam or assessment. You could write:

sqrt(-3+4i) = sqrt(12-22+2x1x2xi)

We write the RHS like that because from:

(a+ib)2=a2-b2+2abi

We take the square root of both sides to obtain:

a+ib= +sqrt(a2-b2+2abi)

Notice how the 1's and 2's correspond to the a's and b's in the RHS, ie a2-b2+2xaxbxi.

One more thing. To find square roots of purely imaginary numbers such as 24i, we write it with the real part too, ie 0+24i and so a2-b2=0 and 2ab=24 and use the above procedure to work out the square root.

If you have any more questions, feel free to ask 