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Question: Write \({\left( {{i^{25}}} \right)^3}\) in polar form....

Write (i25)3{\left( {{i^{25}}} \right)^3} in polar form.

Explanation

Solution

We can simplify the given complex number using the powers of i. We can expand the powers of I and simplify it using the relation i4n+r=ir{i^{4n + r}} = {i^r} . After simplification, we can find its modulus and its argument θ\theta . Then we can express the complex number in the form z=r(cos⁡θ+isin⁡θ)z = r\left( {\cos \theta + i\sin \theta } \right)

Complete step-by-step answer:
Let z=(i25)3z = {\left( {{i^{25}}} \right)^3}
We know that, (ab)c=ab×c{\left( {{a^b}} \right)^c} = {a^{b \times c}}
⇒z=i25×3\Rightarrow z = {i^{25 \times 3}}
On simplification we get,
⇒z=i75\Rightarrow z = {i^{75}}
Now we can write 75 as the product of 4 and remainder.
75=4×18+375 = 4 \times 18 + 3
Using this we get,
⇒z=i4×18+3\Rightarrow z = {i^{4 \times 18 + 3}}
As, ab+c=abac{a^{b + c}} = {a^b}{a^c} and (ab)c=ab×c{\left( {{a^b}} \right)^c} = {a^{b \times c}} , using this we get,
⇒z=(i4)18×i3\Rightarrow z = {\left( {{i^4}} \right)^{18}} \times {i^3}
We know that i4=1{i^4} = 1
⇒z=118×i3\Rightarrow z = {1^{18}} \times {i^3}
Expanding the power of i, we get,
⇒z=i2+1\Rightarrow z = {i^{2 + 1}}
Using ab+c=abac{a^{b + c}} = {a^b}{a^c} , we get,
⇒z=i2×i\Rightarrow z = {i^2} \times i
We know that i2=−1{i^2} = - 1
⇒z=−i\Rightarrow z = - i
Now we have the complex number in the simplest form.
Now we can find its modulus. We know that modulus of a complex number z=x+iyz = x + iy is given by,
r=x2+y2r = \sqrt {{x^2} + {y^2}}
On substituting the values, we get,
⇒r=02+(−1)2\Rightarrow r = \sqrt {{0^2} + {{\left( { - 1} \right)}^2}}
On simplification we get,
⇒r=1\Rightarrow r = 1
We know that argument of a complex number is given by,
θ=tan⁡−1(yx)\theta = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)
On substituting the value, we get,
⇒θ=tan⁡−1(−10)\Rightarrow \theta = {\tan ^{ - 1}}\left( {\dfrac{{ - 1}}{0}} \right)
We know that tan⁡−π2=−10\tan \dfrac{{ - \pi }}{2} = \dfrac{{ - 1}}{0}
⇒θ=−π2\Rightarrow \theta = \dfrac{{ - \pi }}{2}
Now we have the argument and modulus. We know that the complex number in polar form is given by, z=r(cos⁡θ+isin⁡θ)z = r\left( {\cos \theta + i\sin \theta } \right)
On substituting the values, we get,
z=1(cos⁡−π2+isin⁡−π2)z = 1\left( {\cos \dfrac{{ - \pi }}{2} + i\sin \dfrac{{ - \pi }}{2}} \right)
Therefore, the required polar form is z=1(cos⁡−π2+isin⁡−π2)z = 1\left( {\cos \dfrac{{ - \pi }}{2} + i\sin \dfrac{{ - \pi }}{2}} \right).

Note: We know that complex numbers can be plotted in a plane. The complex number z=x+iyz = x + iy is represented in its polar form as z=r(cos⁡θ+isin⁡θ)z = r\left( {\cos \theta + i\sin \theta } \right) where r is the modulus and θ\theta is the argument of the complex number.
Modulus of a complex number is the distance from the origin to the complex number in a plane. It is given by the equation r=x2+y2r = \sqrt {{x^2} + {y^2}} . Argument of the complex number is the angle that the modulus of the complex number makes with the positive x axis. The angle measured in counter-clockwise direction is positive and angle measured in clockwise direction is negative.