z^4=1-i

1-i=sqr(2)[cos(7pi/4)+isin(7pi/4)]

=>

z^4=[2^(1/8)]cis(7pi/4+2kpi)

=>

z=(2^1/8)cis(7pi/16+kpi/2)

{if z^n=r(cosA+isinA), then

z=[r^(1/n)][cos(A/n)+isin(A/n)]

is the De Moivres theorem, &

cisA=cosA+isinA}

=>

the 4 roots are

z0=1.091cis(7pi/16), k=0

z1=1.09 cis(15pi/16), k=1

z2=1.09 cis(pi+7pi/16), k=2

z3=1.09 cis(pi+15pi/16), k=3

From these information, you can

draw the Argand diagrams of the

4 roots.

The roots mean the solutions of the

equation z^4=1-i.

It is found that here z3(your z1) is in

the 1st Q. & z2(your z2) is in the 2nd

Q. Thus

Z2/Z3=cis(pi+7pi/16-pi-15pi/16)

=>

Z2/z3=cis(-pi/2)=cos(-pi/2)+isin(-pi/2)

=>

z2/z3=0-i (a=0, b=-1)

# Math: Complex Numbers?

We're learning about complex numbers and de Moivres theorem and im really confused...

Here is a question:

(a) Use de Moivre’s theorem to find the roots of the equation z^4 = 1 – i.

(b) Draw these roots on an Argand diagram.

(c) If z1 is the root in the first quadrant and z2 is the root in the second quadrant, find z2/ z1 in the form a + ib.

Can you please do this problem and explain how you do it???

What exactly do they mean by roots? i dont get it..

Here is a question:

(a) Use de Moivre’s theorem to find the roots of the equation z^4 = 1 – i.

(b) Draw these roots on an Argand diagram.

(c) If z1 is the root in the first quadrant and z2 is the root in the second quadrant, find z2/ z1 in the form a + ib.

Can you please do this problem and explain how you do it???

What exactly do they mean by roots? i dont get it..

## 2 Answers

- 10
- a root of an equation is a value that satisfies the equation. a root of a function is a value x for which f(x)=0.

roots also refer to the inverse of taking powers. the square root of x are the two values y for which y^2 is x.

The 4th roots of 1-i are the 4 values z for which z^4 = 1-i.

DeMoivre’s Theorem says that one nth root of a complex number z^n is that number in polar form whose modulus is the nth root of the original and whose argument is the original (plus or minus some multiple of 2 pi) divides by n.

1-i is in Q4 on the complex plane. There are 4 complex roots each separated by 2 pi/n or 360°/n.

1-i = 1+(-1)i = a+bi where a = 1, b = -1. Its modulus is square root of (a^2+b^2) = sqrt(2). Its argument is -45° (-pi/4) or 315° (7 pi/4).

Thus the 4th root has modulus (2^(1/2))^(1/4) = 2^(1/8) or the eighth root of 2.

Equivalent angles to 7 pi/4 are

7 pi/4, 15 pi/4, 23 pi/4, 31 pi/4

The corresponding roots have angles

7 pi/16, 15 pi/16, 23 pi/16, 31 pi/16.10

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