complex numbers?

Question

We did the example
Find the solutions of z^2 = 5i

The first line to solve this was
5i = 5(cos pi/2 + sin pi/2)
so z= sqrt5 (cos pi/4 + i sin pi/4)

I cannot see how these lines were obtained. could you help
thanks

the question i am really asking is i dont know why z^2 is equal to 5(cospi/2+sinpi/2)

Answer 1

Find the solutions of z^2 = 5i

5i = 5(cos pi/2 + i*sin pi/2)
simply breaks the complex
number into it's real and
imaginary parts

5 cos pi/2 =5*0=0 (the real
part) and
5i*sine pi/2 =5*i*1=5i (the
imaginary part)
hence,5i = 0+5i

5i = 5(cos pi/2 + isin pi/2)

the sqrt (5i)is
sqrt5(cospi/4+i*sinpi/4)
=sqrt5(1/sqrt2+i*(1/sqrt2))
=sqrt(5/2)+sqrt(5/2)*i

square this answer and you
get your 5i

however,there are two square
roots of 5i in the complex
plane-i^2 times our principal
root that is,
i^2*(sqrt(5/2)+sqrt(5/2)*i)
= -sqrt(5/2)-sqrt(5/2)*i

this is also 5i when squared
the two roots are pi apart

{note; there are three cube roots
and they are 2pi/3 apart}

i hope that this helps

Answer 2

Mond,

Como has answered (half of) the second part but not the first.

Firstly, you need to understand that complex numbers can be written in mod/arg form as well as rectangular (ie x/y) form. For example, a number like 5i can be seen as 0 in the x direction and 5 in the y. But it can also be seen as a distance of 5 from the origin at an angle of pi/2 (90deg) from the positive, real axis.

Once you have found the modulus, r (r>0), and the argument, A, then you can write the complex number as z = r(cosA + isinA) (you should have gone through examples of this).

De Moivre's theorem tells you that z^n = r^n (cosnA + isinnA). However, you have to bear in mind that, in our example, A could also have been 5pi/2 since this is just 2pi more than pi/2. So, when you halve the angle, the new angle will be 5pi/4.

So, there are actually 2 square roots of 5i; sqrt5 (cospi/4 + isinpi/4) is one of them. The other is sqrt5 (cos5pi/4 + isin5pi/4). The two roots are exactly pi apart, angularly, so (or course) one = - the other (exactly as with square roots of real numbers).


Hope this helps.


PS Just noticed that you have missed an i in your question.

5i = 5(cospi/2 + isinpi/2)

Answer 3

Z ² = 5.(cos π/2 + i sin π/2)
Z = √5 (cos π/4 + i sin π/4)

This result comes from a theorem known as De Moivre`s Theorem which basically states that:-

To find the square root , take square root of number outside bracket and divide angle by 2

Answer 4

your 2nd equation is equal to 5i as is your 1st one. This means 5i can be substitued and then square rooted to give z