if z = 2+ i, how 2 find the argument of z^5?

Question

this symbol ( ^ ) means to the power of .
Thanks a lot in advance

Answer 1

(2 + i) * (2 + i) * (2 + i) * (2 + i) * (2 + i)

Multiplication is commutitave, so no order matters. Split it up into smaller problems.

(2 + i) * (2 + i) = 4 + 4i + i^(2)

There are two sets of those, so you have 2(4 + 4i + i^(2))

This equals 8 + 8i + 2i^(2). there is one more factor of (2 + i)

so the final problem is reduced to (8 + 8i +2i^(2))*(2 + i)

Should be easy from there! Hope this helps.

Answer 2

The point 2 + i has argument arctan (1/2). (When looking at a point x + iy, if x is positive, then the argument will be arctan (y/x). )

So the point z^5 has argument 5 arctan (1/2). (When taking the fifth power of a complex number, you take its magnitude to the fifth power, and multiply its argument by 5. We don't care about the magnitude for this problem, though.)

The angle 5 arctan (1/2), it turns out, lies between -π and π, so we do not have to worry about moving outside the principal branch of the argument. (That is, we do not need to add or subtract 2π to get a value in the usual range.)

So the answer is 5 arctan (1/2), which is about 2.31823805.

Answer 3

Hey there!

In order to find the arguement of complex number a+bi, use the following formula.

θ=arctan(b/a), where θ is the argument.

Let's use the formula, in order to find the argument of 2+i.

θ=arctan(b/a) --> Write the formula.
θ=arctan(1/2) Substitute 1 for b and 2 for a.

If we try to evaluate the inverse tangent of 1/2, we have to approximate the answer. Let θ=u, where u=arctan(1/2).

Let's first find r. Recall that r is the modulus of complex number a+bi. In order to find r, use the Pythagorean Theorem.

r=sqrt(a^2+b^2) --> Write the formula.
r=sqrt(2^2+1^2) --> Substitute 2 for a and 1 for b.
r=sqrt(4+1) --> Square 2 and square 1, respectively.
r=sqrt(5) Add 4 and 1.

So r=sqrt(5). The trigonemtric form of the complex number, a+bi, is.

a+bi=r(cos(θ)+isin(θ)), where r is the modulus and θ is the argument.

If we let n be the exponent on both sides of the equation, the formula will look like this.

(a+bi)^n=(r(cos(θ)+isin(θ)))^n.

Let's use De Moivre's Theorem for powers, on the right side of the equation. The formula would look like this.

(a+bi)^n=r^n(cos(nθ)+isin(nθ)).

Let's substitute 2 for a, 1 for i, 5 for n, sqrt(5) for r and u for θ. The answer would look like this.

(2+i)^5=(sqrt(5))^5(cos(5u)+isin(5u)).

You can rewrite (sqrt(5))^5 as 25sqrt(5).

(2+i)^5=25sqrt(5)(cos(5u)+isin(5u)).

Substituting arctan(1/2), the answer would now be.

(2+i)^5=25sqrt(5)(cos(5(arctan(1/2))+isin(5(arctan(1/2))).

So the arugment of the answer is 5(arctan(1/2)). If you want to approximate the answer, the answer would be about 2.32.

If you want to know the answer, evaluate, 25sqrt(5)(cos(5arctan(1/2)+isin(5arctan(1/2))), on your calculator. The answer would then be -38+41i.

Hope it helps!

Answer 4

If z = 2+ i,

Argument of z^5 is 5 times the argument of z.

Argument of z is tan inverse of 1/2 = 0.4637 radian.

Argument of z^5 is 5 times of 0.4637 = 2.3185 radian

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If you compute the value of z^5, you will get

z^5 = -38 + 41i

Argument of z^5 is tan inverse of - 41/38 = - 0.82335 radian.

Note that 2.3185 radian is [π - 0.82335] = 2.3185 radian.
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Therefore argument of z^5 is 5 times argument of z

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Answer 5

z^5 = (2 + i)^5
Remember i*i = -1

(2 + i)(2 + i) = 2*2 + 2*i + 2*i + i*i = 3 + 4i
(3 + 4i)(2 + i) = 6 + 8i + 3i - 4 = 2 + 11i
(2 + 11i)(2 + i) = 4 + 22i + 2i - 11 = -7 + 24i
(-7 + 24i)(2 + i) = -14 - 7i +48i -24 = -38 +41i
z^5 = -38 + 41i

Answer 6

You don't have to compute z^5 to do this.
The argument of 2+i is arctan(1/2).
The argument of (2+i)^5 is therefore
5*arctan(1/2).
The general idea is to write z as r* e^(it) = r(cos t + i sin t.)
Then z^n = r^n *e^(nit) = r^n( cos nt + i sin nt).
So you can see that if the argument of z is t
then the argument of z^n is nt.

Answer 7

(2+i)^2=2^2+2.2.i+i^2=4+4i+i^2=4+4i-1=3-4i
(3-4i)^2=9-12i+16i^2=9-12i-16=-7-12i
(2+i) (-7-12i)=-14-24i-7i-12i^2=-14-31i+12=-2-31i

Answer 8

(2+i)^2=2^2+2.2.i+i^2=4+4i+i^2...
(3-4i)^2=9-12i+16i^2=9-12i-16=...
(2+i) (-7-12i)=-14-24i-7i-12i^2=-14-...