part1. Let z=rcisθ where 0<θ<pi. show that Arg(z^2)=2θ. part2. sketch the region defined by {z: Arg(z^2)>pi/2} in the answers it shows region pi/4 to pi/2 and region 5pi/8...

part1. Let z=rcisθ where 0<θ<pi. show that Arg(z^2)=2θ. part2. sketch the region defined by {z: Arg(z^2)>pi/2} in the answers it shows region pi/4 to pi/2 and region 5pi/8...

i that right????????

imaginary numbers. complex numbers.radicals. order in a solution

How do you put the complex number -2+i into a form of e^{ix} ? with i = sqrt(-1) x= angle...

So generally 2 the power of 0 equals 1. same goes for 3 and 4 and so on. You know what I mean. So what is zero in the imaginary dimesion? i is the first of the imaginry numbers, like 1 is in the natural...

8+(x+2y)i=x+2i I need to find the value of x and y, assuming they are both real numbers. How do I go about solving this? Thanks!

Is 3i the fifth root of -243i? Justify please.

Find the indicated power using De Moivre's Theorem. (Express your fully simplified answer in the form a + bi.) ((sqrt)3 - i)^6

What is the relation between the set of complex numbers, the imaginary numbers, and the real numbers?

(x+yi)4=-7-24I Find x and y In complex numbers

3x^3+4x^2−7x+2=0 x^4+8x^3+6x^2−5x+14=0 a polynomial has real coefficients. the degree is 4. two zeros are i and 9+i.

the above is what I got from (4-2i)/(-10+5i) multiplied by (3-i)/(-3+i) written as equation/equation x equation/equation thank you for any help

it is apart of 3i+[1/2]-[1/3](1-3i)+2

(3i)(6i^2) ( have to simplifiy) Not good with imaginary numbers:( Thanks again

Evaluate these powers of i.If there is an imaginary part,be sure to enter your answer in a+bi form,where both a and b are real numbers i1113 i-147

9x^4-28x^2+3=0

Hi guys I had a quick question? How do you determine the minimum degree of a function. Do you look at the roots or the extrema in the graph? If you do look at the x intercepts couldn't that answer...

(2+i)(-3-3i) and ...

n = 3 -4 and 2i are zeros f(-1) = -45 Find the expanded and simplified polynomial

How would you solve the expressions 2+3i and 2-3i where I is an imaginary number? (i=√-1)