Putting complex numbers into exponential form.?

Question

Put the complex number 2 + 3i into its exponential form and hence find the three

complex roots of

z^3- 2 - 3i = 0

Determine the Cartesian form of these three roots.

Answer 1

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Answer 2

it would help by studying your math book

i will give you a link on euler's formula, this will help you understand

http://en.wikipedia.org/wiki/Euler%27s_f...

Answer 3

2 + 3 i = √13 /_ Ó¨  where Ó¨ = tan ^(-1) (3/2) = 0.625

Z³ = √13 [cos (0.625 + 2kπ) + i sin (0.625 + 2kπ ) ]

Z is , for k = 0 , 1 , 2 , given by :-

13^(1/6) [cos {(0.625 + 2kπ)/3}+ i sin {(0.625 + 2kπ)/3]

z1 = 13^(1/6) ( cos 0.208 + i sin 0.208 )

z2 = 13^(1/6) ( cos 2.3+ i sin 2.3)

Z3 = 13^(1/6) ( cos 4.4+ i sin 4.4)

z1 = 13^(1/6) ( 0.98 + i 0.21 )

z2 = 13^(1/6) ( - 0.67 + i 0.75)

Z3 = 13^(1/6) (- 0.31- i 0.95)

Answer 4

2+3i = (rt13)[ cosT +isinT] where cosT = 2/rt13 and sinT= 3/rt13

2+3i = (rt13) e^(iT) etc.

So z = [(rt13) e^(iT)]^(1/3)

        = (6th root of 13) {Cos[(2pik/3)+(T/3)] +isin[(2pi/3)+(T/3)]} for k = 0,1 and 2.

Answer 5

Let x=2 and y=3

r=sqrt[4+9]=sqrt[13]

theta=tan[3/2]

The exponential form of 2+3i is sqrt[13]*exp[tan(3/2)]

Now, z^3=2+3i=sqrt[13]*exp[tan(3/2)]=

sqrt[13]*exp[2npi+tan(3/2)]

==>z=sixth root[13]*exp[tan(3/2)] or

sixthroot[13]*exp[2pi+tan(3/2)]

or sixthroot[13]*exp[4pi+tan(3/2)]