Chapter 9: Q. 8 (page 591)

If $z = r (\cos θ + i \sin θ)$ is a complex number, then
$z^{n} =_[\cos () + i \sin ()]$ .

Short Answer

Expert verified

The general solution is

$z^{n} = r^{n} [\cos (n θ) + i \sin (n θ)]$

Step by step solution

Step 1. Given

Given complex number in polar form

$z = r \cos θ + i \sin θ$

Calculation

$Z_{1} Z_{2} = r_{1} (\cos θ_{1} + i \sin θ_{1}) \times r_{2} (\cos θ_{2} + i \sin θ_{2}) = r_{1} r_{2} (\cos θ_{1} \cos θ_{2} + i \cos θ_{1} \sin θ_{2} + i \sin θ_{1} \cos θ_{2} + i^{2} \sin θ_{1} \sin θ_{2}) = r_{1} r_{2} [(\cos (θ_{1} + θ_{2}) + i \sin (θ_{1} + θ_{2})] S i m i l a r y f o r n t i m e s a g e n e r a l i s e s o l u t i o n w i l l b e z^{n} = r^{n} [\cos (n θ) + i \sin (n θ)]$

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If $z = r (\cos θ + i \sin θ)$ is a complex number, then
$z^{n} =_[\cos () + i \sin ()]$ .

Short Answer

Step by step solution

Step 1. Given

Calculation

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If z=r(cosθ+isinθ)is a complex number, thenzn=_____[cos(____)+isin(____)].

Short Answer

Step by step solution

Step 1. Given

Calculation

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If $z = r (\cos θ + i \sin θ)$ is a complex number, then
$z^{n} =_[\cos () + i \sin ()]$ .