Complex numbers?

Question

Calculate values of complex roots for (-1+i) to the power of 1/3

Express hyp sine expession of sinh (2i) in form a+ib where a and b are real

Answer 1

Whenever working with roots of complex numbers, it is easiest to use exponential notation. Meaning:

a + ib = r*e^(i t)

Where r = sqrt(a^2 + b^2) and tan(t) = b/a

In this form, complex roots are easy because:

(r * e^(i t))^(1/n) = r^(1/n) * e^(i t/n)

As a final step, you apply the Euler identity: e^(i t) = cos(t) + i sin(t)

The multiple roots arise from the cyclic nature of e^(i t) because e ^ (i t) = e^(i (t + 2m π)) for all integer m. This means there are n distinct nth roots. This is true for real numbers as well. for example, the cube roots of 1 are 1, (sqrt(3)/2 + i/2), (sqrt(3)/2 - i/2). You can multiply these out to see they are really roots.

As for your problem: (-1 + i)^(1/3)

-1 + i = 2^(1/2) * e^ (i 3π/4)

So the roots are 2^(1/6) e^(i (2m π + 3π/4)/3) for m = 0, 1, 2.

For example when m=1, the root is 2^(1/6) e^(i (11π/12))

Using the euler identity, this is:

2^(1/6)(cos(11π/12) + i sin(11π/12))

Using some obscure trig identities, you can reduce this to:

2^(-4/3) * (-sqrt(3) - 1 + i (sqrt(3) - 1))

Your second question on the sinh(2i) just requires using the idetities of the sinh and sine:

sinh(x) = (e^x - e^(-x))/2 and sin(x) = (e^(ix) - e^(-ix))/2i

From this it is obvious that sinh(ix) = i sinx so sinh(2i) = 0 + i sin(2).

Answer 2

The sq. root of an imaginary extensive type is yet another imaginary extensive type. First be conscious that (a+bi)^2 = a^2 +2abi -b^2 = a^2-b^2 + (2ab)i So working example, (2+i)^2 = 4 +4i -a million) =3+4i consequently the sqrt of three+4i is two+i and 2-a million on account that complicated numbers consistently are available conjugate pairs. So enable's see a thank you to teach that the sqrt of three+4i is unquestionably 2 + i and 2 - i bear in mind (a+bi)^2= a^2 -b^2 + (2ab)i So we ought to teach that a^2 - b^2 = 3 and 2ab = 4 positioned b= 2/.a into first equation getting a^2-4/a^4=3 a^4 -3a^2 -4 =0This is a quadratic in a^2 which would be resolve by factoring giving a^2 = 4 or -a million. Reject the -a million as we'd like a real answer so x^2 = 4 and a = 2 or -2 b=2/a so b= a million or -a million for this reason the sqrt of three+=4i is two+i or 2-i'm hoping this helped.

Answer 3

To find the roots of a complex number it is normally easiest to write it in polar form: r*e^{i theta} = r*(cos(theta) + i sin(theta)).

To do this with -1 + i first you need to find r. r^2 = (-1+i)*(-1-i) = 2 so r = sqrt(2). So, dividing by sqrt of two you have

cos(theta) + i sin(theta) = -1/sqrt(2) + i/sqrt(2).

Thus theta = 3pi/4. So (-1 + i) = sqrt(2) e^{i 3pi/4}. Now, there are three cube roots of this number and the come by multiplying one of the roots by cube roots of one. It's easy to find one root of (-1+i) by taking the cube root of sqrt(2) and dividing theta by 3. Doing this you get

sixthroot(2) e^{i pi/4}.

Now the cube roots of 1 are 1, e^{i 2pi/3}, and e^{i 4pi/3}. Thus the three cube roots of (-1 + i) are

sixthroot(2) e^{i pi/4}, sixthroot(2) e^{i 11pi/12}, sixthroot(2) e^{i 19pi/12}.

To solve the other problem use the formulas sinh(theta) = (e^theta - e^-theta)/2 and e^{i theta} = cos(theta) + i sin(theta).