Polar form of Complex Number

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Polar form

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Let \(z=x+iy\) be a complex number with magnitude r and amplitude \( \theta \) , then
\( r=| z |=\sqrt{x^2 +y^2} \) , and \( \theta \tan^{-1}( \frac{y}{x} )\) Using right angle triangle, we get that
\( \cos \theta=\frac{b}{h}\)
or \( \cos \theta=\frac{x}{r}\)
or \( x= r \cos \theta \)
Similarly,
\( \sin \theta=\frac{p}{h}\)
or \( \sin \theta=\frac{y}{r}\)
or \( y= r \sin \theta \)
Hence, polar form of complex number is
\( z=x+iy\)
or \( z= r\cos \theta +ir \sin \theta \)
or \( z=r( \cos \theta +i\sin \theta )\)
NOTE
A complex number z=x+iy is represented by polar coordinate \( (r,\theta)\) , as
\( z=r \cos \theta +i\sin \theta \)

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Worked out Examples: polar form

Express the followig complex numbers in trigonometric (polar) forms.

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1. -1
   

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   Solution,
   Given that,
   z = -1
   or z = -1 + 0i
   Comparing -1 + 0i with x + iy, We get
   
   x = -1 and y = 0
   Then, the magnitude of z is given by
   
   \( r = \sqrt{x^2 + y^2} \)
   or \(r = \sqrt{(-1)^2 + (0)^2}\)
   so r = 1
   Again, the amplitude of z is giveb by
   \( \theta = \tan^{-1} (\frac{y}{x} )\)
   or \( \theta = \tan^{-1}( \frac{0}{-1} )\)
   or \( \theta = -\tan^{-1} (\frac{0}{1}) \)
   or \( \theta = -\tan^{-1} (0) \)
   According to quadrant rule, (-,+) belong to second quadrant

   (-,+)→(II) ✅	❌(I)←(+,+)
   (-,-)→(III)❌	❌(IV)←(+,-)

   Therefore
   θ = 180
   Therefore, the polar form of z is
   z = r(cosθ + isinθ)
   or z = cos180 + isin180
2. z=-5+5i
   

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   Solution,
   Given that,
   z =-5+5i
   or z = 5(-1+1i)
   Comparing 5(-1+1i) with x + iy, We get
   x = -1 and y = 1
   Then, the magnitude of z is given by
   
   \( r = 5 \sqrt{(x)^2 + (y)^2} \)
   or \(r = 5 \sqrt{(-1)^2 + (1)^2}\)
   so \(r = 5 \sqrt{2}\)
   Again, the amplitude of z is giveb by
   \( \theta = \tan^{-1} (\frac{y}{x} )\)
   or \( \theta = \tan^{-1}( \frac{1}{-1} )\)
   or \( \theta = -\tan^{-1} (\frac{1}{1}) \)
   or \( \theta = -\tan^{-1} (1) \)
   According to quadrant rule, (-,+) belong to second quadrant

   (-,+)→(II) ✅	❌(I)←(+,+)
   (-,-)→(III)❌	❌(IV)←(+,-)

   Therefore
   θ = 135
   Therefore, the polar form of z is
   z = r(cosθ + isinθ)
   or z = cos135 + i sin135
   or\( z=5\sqrt{2}( \cos \frac{3\pi }{4}+\sin \frac{3\pi }{4}i )\)
3. z=3i
   

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   Solution,
   Given that,
   z = 3i
   or z = 3(0 + i)
   Comparing 3(0 + i) with x + iy, We get
   x = 0 and y = 1
   Then, the magnitude of z is given by
   
   \( r = 3 \sqrt{(x)^2 + (y)^2} \)
   or \(r = 3 \sqrt{(0)^2 + (1)^2}\)
   so \(r = 3\)
   Again, the amplitude of z is given by
   \( \theta = \tan^{-1} \left( \frac{y}{x} \right) \)
   Since \( x = 0 \), the amplitude θ directly corresponds to:
   \( \theta = \frac{\pi}{2} \) or \( \theta = 90^\circ \)
   Therefore,
   θ = 90°
   Therefore, the polar form of z is
   z = r(cosθ + isinθ)
   or z = 3(cos90° + i sin90°)
   
4. z = 1 + i
   

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   Solution,
   Given that,
   z = 1 + i
   or z = 1(1 + i)
   Comparing 1(1 + i) with x + iy, We get
   x = 1 and y = 1
   Then, the magnitude of z is given by
   
   \( r = \sqrt{(x)^2 + (y)^2} \)
   or \(r = \sqrt{(1)^2 + (1)^2}\)
   so \(r = \sqrt{2}\)
   Again, the amplitude of z is given by
   \( \theta = \tan^{-1} \left( \frac{y}{x} \right) \)
   or \( \theta = \tan^{-1} \left( \frac{1}{1} \right) \)
   or \( \theta = \tan^{-1} (1) \)
   According to quadrant rule, (+,+) belong to first quadrant
   

   (-,+)→(II) ❌	✅(I)←(+,+)
   (-,-)→(III)❌	❌(IV)←(+,-)

   Therefore,
   θ = 45°
   Therefore, the polar form of z is
   z = r(cosθ + isinθ)
   or z = √2(cos45° + i sin45°)
   
5. z = 1 - i
   

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   Solution,
   Given that,
   z = 1 - i
   or z = 1(1 - i)
   Comparing 1(1 - i) with x + iy, We get
   x = 1 and y = -1
   Then, the magnitude of z is given by
   
   \( r = \sqrt{(x)^2 + (y)^2} \)
   or \(r = \sqrt{(1)^2 + (-1)^2}\)
   so \(r = \sqrt{2}\)
   Again, the amplitude of z is given by
   \( \theta = \tan^{-1} \left( \frac{y}{x} \right) \)
   or \( \theta = \tan^{-1} \left( \frac{-1}{1} \right) \)
   or \( \theta = \tan^{-1} (-1) \)
   According to quadrant rule, (+,-) belong to fourth quadrant
   

   (-,+)→(II) ❌	❌(I)←(+,+)
   (-,-)→(III)❌	✅(IV)←(+,-)

   Therefore,
   θ = 315°
   Therefore, the polar form of z is
   z = r(cosθ + isinθ)
   or z = √2(cos(315°) + i sin(315°))
   
6. z = -1 - i
   

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   Solution,
   Given that,
   z = -1 - i
   or z = 1(-1 - i)
   Comparing 1(-1 - i) with x + iy, We get
   x = -1 and y = -1
   Then, the magnitude of z is given by
   
   \( r = \sqrt{(x)^2 + (y)^2} \)
   or \(r = \sqrt{(-1)^2 + (-1)^2}\)
   so \(r = \sqrt{2}\)
   Again, the amplitude of z is given by
   \( \theta = \tan^{-1} \left( \frac{y}{x} \right) \)
   or \( \theta = \tan^{-1} \left( \frac{-1}{-1} \right) \)
   or \( \theta = \tan^{-1} (1) \)
   According to quadrant rule, (-,-) belong to third quadrant
   

   (-,+)→(II) ❌	❌(I)←(+,+)
   (-,-)→(III)✅	❌(IV)←(+,-)

   Therefore,
   θ = 180° + 45° = 225°
   Therefore, the polar form of z is
   z = r(cosθ + isinθ)
   or z = √2(cos225° + i sin225°)
   
7. z = -1 + i
   

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   Solution,
   Given that,
   z = -1 + i
   or z = 1(-1 + i)
   Comparing 1(-1 + i) with x + iy, We get
   x = -1 and y = 1
   Then, the magnitude of z is given by
   
   \( r = \sqrt{(x)^2 + (y)^2} \)
   or \(r = \sqrt{(-1)^2 + (1)^2}\)
   so \(r = \sqrt{2}\)
   Again, the amplitude of z is given by
   \( \theta = \tan^{-1} \left( \frac{y}{x} \right) \)
   or \( \theta = \tan^{-1} \left( \frac{1}{-1} \right) \)
   or \( \theta = \tan^{-1} (-1) \)
   According to quadrant rule, (-,+) belong to second quadrant
   

   (-,+)→(II) ✅	❌(I)←(+,+)
   (-,-)→(III)❌	❌(IV)←(+,-)

   Therefore,
   θ = 180° - 45° = 135°
   Therefore, the polar form of z is
   z = r(cosθ + isinθ)
   or z = √2(cos135° + i sin135°)
   
8. z = -1 - √3i
   

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   Solution,
   Given that,
   z = -1 - √3i
   or z = 1(-1 - √3i)
   Comparing 1(-1 - √3i) with x + iy, We get
   x = -1 and y = -√3
   Then, the magnitude of z is given by
   
   \( r = \sqrt{(x)^2 + (y)^2} \)
   or \(r = \sqrt{(-1)^2 + (-√3)^2}\)
   so \(r = \sqrt{1 + 3} = \sqrt{4} = 2\)
   Again, the amplitude of z is given by
   \( \theta = \tan^{-1} \left( \frac{y}{x} \right) \)
   or \( \theta = \tan^{-1} \left( \frac{-√3}{-1} \right) \)
   or \( \theta = \tan^{-1} (√3) \)
   According to quadrant rule, (-,-) belong to third quadrant
   

   (-,+)→(II) ❌	❌(I)←(+,+)
   (-,-)→(III)✅	❌(IV)←(+,-)

   Therefore,
   θ = 180° + 60° = 240°
   Therefore, the polar form of z is
   z = r(cosθ + isinθ)
   or z = 2(cos240° + i sin240°)
   
9. z = -1 + √3i
   

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   Solution,
   Given that,
   z = -1 + √3i
   or z = 1(-1 + √3i)
   Comparing 1(-1 + √3i) with x + iy, We get
   x = -1 and y = √3
   Then, the magnitude of z is given by
   
   \( r = \sqrt{(x)^2 + (y)^2} \)
   or \(r = \sqrt{(-1)^2 + (√3)^2}\)
   so \(r = \sqrt{1 + 3} = \sqrt{4} = 2\)
   Again, the amplitude of z is given by
   \( \theta = \tan^{-1} \left( \frac{y}{x} \right) \)
   or \( \theta = \tan^{-1} \left( \frac{√3}{-1} \right) \)
   or \( \theta = \tan^{-1} (-√3) \)
   According to quadrant rule, (-,+) belong to second quadrant
   

   (-,+)→(II) ✅	❌(I)←(+,+)
   (-,-)→(III)❌	❌(IV)←(+,-)

   Therefore,
   θ = 180° - 60° = 120°
   Therefore, the polar form of z is
   z = r(cosθ + isinθ)
   or z = 2(cos120° + i sin120°)
   
10. z = -2√3 + 2i
    

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    Solution,
    Given that,
    z = -2√3 + 2i
    or z = 2(-√3 + i)
    Comparing 2(-√3 + i) with x + iy, We get
    x = -√3 and y = 1
    Then, the magnitude of z is given by
    
    \( r = 2 \times \sqrt{(x)^2 + (y)^2} \)
    or \(r = 2 \times \sqrt{(-√3)^2 + (1)^2}\)
    so \(r = 2 \times \sqrt{3 + 1} = 2 \times \sqrt{4} = 2 \times 2 =4\)
    Again, the amplitude of z is given by
    \( \theta = \tan^{-1} \left( \frac{y}{x} \right) \)
    or \( \theta = \tan^{-1} \left( \frac{1}{-√3} \right) \)
    or \( \theta = \tan^{-1} \left( -\frac{1}{√3} \right) \)
    According to quadrant rule, (-,+) belong to second quadrant
    

    (-,+)→(II) ✅	❌(I)←(+,+)
    (-,-)→(III)❌	❌(IV)←(+,-)

    Therefore,
    θ ≈ 180° - 30° = 150°
    Therefore, the polar form of z is
    z = r(cosθ + isinθ)
    or z = 4(cos150° + i sin150°)
    
11. z = -2 + 2√3i
    

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    Solution,
    Given that,
    z = -2 + 2√3i
    or z = 2(-1 + √3i)
    Comparing 2(-1 + √3i) with x + iy, We get
    x = -1 and y = √3
    Then, the magnitude of z is given by
    
    \( r = 2 \times \sqrt{(x)^2 + (y)^2} \)
    or \(r = 2 \times \sqrt{(-1)^2 + (√3)^2}\)
    so \(r = 2 \times \sqrt{1 + 3} = 2 \times \sqrt{4} = 2 \times 2 = 4\)
    Again, the amplitude of z is given by
    \( \theta = \tan^{-1} \left( \frac{y}{x} \right) \)
    or \( \theta = \tan^{-1} \left( \frac{√3}{-1} \right) \)
    or \( \theta = \tan^{-1} \left( -√3 \right) \)
    According to quadrant rule, (-,+) belong to second quadrant
    

    (-,+)→(II) ✅	❌(I)←(+,+)
    (-,-)→(III)❌	❌(IV)←(+,-)

    Therefore,
    θ ≈ 180° - 60° = 120°
    Therefore, the polar form of z is
    z = r(cosθ + isinθ)
    or z = 4(cos120° + i sin120°)
    
12. z = -3 - √3i
    

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    Solution,
    Given that,
    z = -3 - √3i
    or z = (-3) + (-√3)i
    Comparing (-3) + (-√3)i with x + iy, We get
    x = -3 and y = -√3
    Then, the magnitude of z is given by
    
    \( r = \sqrt{(x)^2 + (y)^2} \)
    or \(r = \sqrt{(-3)^2 + (-√3)^2}\)
    so \(r = \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3}\)
    Again, the amplitude of z is given by
    \( \theta = \tan^{-1} \left( \frac{y}{x} \right) \)
    or \( \theta = \tan^{-1} \left( \frac{-√3}{-3} \right) \)
    or \( \theta = \tan^{-1} \left( \frac{1}{√3} \right) \)
    According to quadrant rule, (-,-) belong to third quadrant
    

    (-,+)→(II) ❌	❌(I)←(+,+)
    (-,-)→(III)✅	❌(IV)←(+,-)

    Therefore,
    \( \theta = 180° + 30° = 210° \)
    Therefore, the polar form of z is
    \( z = r(\cos \theta + i \sin \theta) \)
    or \( z = 2\sqrt{3}(\cos 210° + i \sin 210°) \)
    
13. z = √3 - 3i
    

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    Solution,
    Given that,
    z = √3 - 3i
    or z = √3 + (-3)i
    Comparing √3 + (-3)i with x + iy, We get
    x = √3 and y = -3
    Then, the magnitude of z is given by
    
    \( r = \sqrt{(x)^2 + (y)^2} \)
    or \(r = \sqrt{(\sqrt{3})^2 + (-3)^2}\)
    so \(r = \sqrt{3 + 9} = \sqrt{12} = 2\sqrt{3}\)
    Again, the amplitude of z is given by
    \( \theta = \tan^{-1} \left( \frac{y}{x} \right) \)
    or \( \theta = \tan^{-1} \left( \frac{-3}{\sqrt{3}} \right) \)
    or \( \theta = \tan^{-1} \left( -\sqrt{3} \right) \)
    According to quadrant rule, (+, -) belong to fourth quadrant
    

    (-,+)→(II) ❌	❌(I)←(+,+)
    (-,-)→(III)❌	✅(IV)←(+,-)

    Therefore,
    \( \theta = 360° - 60° = 300° \)
    Therefore, the polar form of z is
    \( z = r(\cos \theta + i \sin \theta) \)
    or \( z = 2\sqrt{3}(\cos 300° + i \sin 300°) \)
    

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Worked out Example: Standard form

Express the following in standard form: Cartesian form.

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1. \( \sqrt{2} (\cos 135° + i \sin 135°) \)
   

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   Given that,
   \(z= \sqrt{2} (\cos 135° + i \sin 135°) \)
   We know:
   \( \cos 135° = -\frac{1}{\sqrt{2}} \)
   \( \sin 135° =\frac{1}{\sqrt{2}} \)
   Substituting these values into the polar form, we get
   \( z = \sqrt{2} \left( -\frac{1}{\sqrt{2}} + i \cdot \frac{1}{\sqrt{2}} \right) \)
   or\( z = -1 + i \)
   Therefore, the standard form (rectangular form) of \( \sqrt{2} (\cos 135° + i \sin 135°) \) is \( -1 + i \).
   
2. \( 5 \sqrt{2} (\cos 315° + i \sin 315°) \)
   

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   Solution
   Given that,
   \( z = 5 \sqrt{2} (\cos 315° + i \sin 315°) \)
   We know:
   \( \cos 315° = \frac{1}{\sqrt{2}} \)
   \( \sin 315° = \frac{-1}{\sqrt{2}} \)
   Substituting these values into the polar form, we get
   \( z = 5 \sqrt{2} \left( \frac{1}{\sqrt{2}} + i \cdot \frac{-1}{\sqrt{2}} \right) \)
   or \( z = 5 - 5i \)
   Therefore, the standard form (rectangular form) of \( 5 \sqrt{2} (\cos 315° + i \sin 315°) \) is \( 5 - 5i \).
   
3. \( 6 (\cos \frac{4 \pi}{3} + i \sin \frac{4 \pi}{3}) \)
   

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   Solution,
   Given that,
   \( z = 6 (\cos \frac{4 \pi}{3} + i \sin \frac{4 \pi}{3}) \)
   We know:
   \( \cos \frac{4 \pi}{3} = -\frac{1}{2} \)
   \( \sin \frac{4 \pi}{3} = -\frac{\sqrt{3}}{2} \)
   Substituting these values into the polar form, we get
   \( z = 6 \left( -\frac{1}{2} - i \cdot \frac{\sqrt{3}}{2} \right ) \)
   or\( z = -3 - 3i \)
   Therefore, the standard form (rectangular form) of \( 6 (\cos \frac{4 \pi}{3} + i \sin \frac{4 \pi}{3}) \) is \( -3 - 3i \).
   
4. \( 6 (\cos \frac{5 \pi}{3} + i \sin \frac{5 \pi}{3}) \)
   Solution,
   

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   Given that,
   \( z = 6 (\cos \frac{5 \pi}{3} + i \sin \frac{5 \pi}{3}) \)
   We know:
   \( \cos \frac{5 \pi}{3} = \frac{1}{2} \)
   \( \sin \frac{5 \pi}{3} = -\frac{\sqrt{3}}{2} \)
   Substituting these values into the polar form, we get
   \( z = 6 \left( \frac{1}{2} + i \cdot \left(-\frac{\sqrt{3}}{2}\right) \right) \)
   or \( z = 3 - 3i \)
   Therefore, the standard form (rectangular form) of \( 6 (\cos \frac{5 \pi}{3} + i \sin \frac{5 \pi}{3}) \) is \( 3 - 3i \).
   
5. \( 7 (\cos \pi + i \sin \pi) \)
   Solution,
   

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   Given that,
   \( z = 7 (\cos \pi + i \sin \pi) \)
   We know:
   \( \cos \pi = -1 \)
   \( \sin \pi = 0 \)
   Substituting these values into the polar form, we get
   \( z = 7 \left( -1 + i \cdot 0 \right) \)
   Simplifying further,
   \( z = 7 \cdot (-1) \)
   Therefore, the standard form (rectangular form) of \( 7 (\cos \pi + i \sin \pi) \) is \( -7 \).
   

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The essence of Polar form

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The relevance of complex number in polar form is that multiplication and division are simpler with this form than the Cartesian form.

Let \( z_1 =r_1 (\cos \theta_1 +i\sin \theta_1 )\) and \( z_2 =r_2 (\cos \theta_2 +i\sin \theta_2 )\) be two complex numbers, then
\( z_1 z_2 =r_1 r_2 (\cos (\theta_1 +\theta_2 )+i\sin (\theta_1 +\theta_2 ))\)
Proof
Given that \( z_1 =r_1 (\cos \theta_1 +i\sin \theta_1 )\) and \( z_2 =r_2 (\cos \theta_2 +i\sin \theta_2 )\) . Thus
\( z_1 z_2 = r_1 (\cos \theta_1 +i\sin \theta_1 ) \times r_2 (\cos \theta_2 +i\sin \theta_2 ) \)
or \( z_1 z_2 = r_1 r_2 (\cos \theta_1 +i\sin \theta_1 )(\cos \theta_2 +i\sin \theta_2 )\)
or \( z_1 z_2 = r_1 r_2 [\cos \theta_1 (\cos \theta_2 +i\sin \theta_2 )+i\sin \theta_1 (\cos \theta_2 +i\sin \theta_2 ) ]\)
or \( z_1 z_2 = r_1 r_2 [\cos \theta_1 \cos \theta_2 +i\cos \theta_1\sin \theta_2 +i\sin \theta_1\cos \theta_2 -\cos \theta_1\sin \theta_2 ]\)
or \( z_1 z_2 = r_1 r_2 [(\cos \theta_1 \cos \theta_2-\cos \theta_1\sin \theta_2 ) +i(\cos \theta_1\sin \theta_2 +\sin \theta_1\cos \theta_2) ]\)
or \( z_1 z_2 = r_1 r_2 [\cos (\theta_1 +\theta_2) +i \sin (\theta_1+ \theta_2)]\)

Theorem

\( arg( z_1 z_2 )=arg z_1 +argz_2 \)

Proof

Let \( z_1 =r_1 ( \cos \theta_1 +i \sin \theta_1 )\) and \( z_2 =r_2 ( \cos \theta_2 +i\sin \theta_2 )\) then
\( z_1 .z_2=r_1 .r_2 [ \cos ( \theta_1 +\theta_2 )+i\sin ( \theta_1 +\theta_2 ) ]\)
Thus,
\( arg( z_1 z_2 )=argz_1 +argz_2 \)

Note:
Any complex number z has infinite arguments; all differ by multiple of \( 2\pi \) . The principal value is in the interval \( [ -\pi ,\pi ]\)

Some important property

1. Let \( z_1 \text{ and } z_2 \) be two complex number then \( arg( z_1 .z_2 )=argz_1 +argz_2 \)
   The argument of product of two complex number is sum of their arguments.
   Proof
   Let \( z_1 =r_1 e^{ i\theta_1}\) and \( z_2 =r_2 e^{ i\theta_2}\) then
   \( z_1 z_2 =( r_1 e^{ i\theta_1} )( r_2 e^{ i\theta_2} )\)
   or \( z_1 z_2 =( r_1 r_2 )e^{i( \theta_1 +\theta_2 )}\)
   Hence,
   \( arg( z_1 .z_2 )=\theta_1 +\theta_2 =argz_1 +argz_2 \)
2. Let\( z_1 \text{ and }z_2 \) be two complex number then,
   \( arg( \frac{z_1 }{z_2 } )=argz_1 -argz_2 \)
   
   The argument of quotient of two complex number is difference of their arguments.
3. Argument of complex number of the form \( z=a+i.0, a > 0\) is 0
4. Argument of complex number of the form \( z=a+i.0, a < 0\) is \( \pi \)
5. Argument of complex number of the form \( z=0+i.b, b > 0\) is \( \frac{\pi }{2}\)
6. Argument of complex number of the form \( z=0+i.b, b < 0\) is \( -\frac{\pi }{2}\)

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If \( z =r( \cos \theta +i\sin \theta )\) then show that \( z^{-1} =\frac{1}{r}( \cos \theta -i\sin \theta )\)

Question

Let \( z=r( \cos \theta +i\sin \theta )\) be a complex number and its inverse is \( z^{-1} =R( \cos \phi +i\sin \phi )\) then
\( zz^{-1} =1\)
or \( r( \cos \theta +i\sin \theta )R( \cos \phi +i\sin \phi )=( 1,0 )\)
or \( rR\{ \cos ( \theta +\phi )+i\sin ( \theta +\phi ) \} =( 1,0 )\)
or \( rR\cos ( \theta +\phi )=1 \text{ and } rR\sin ( \theta +\phi )=0\)
Here
\( rR\sin ( \theta +\phi )=0\)
or \(\sin ( \theta +\phi )=0\)
or \( \theta +\phi =0\)
or \( \theta = - \phi \)
So,
\( rR\cos ( \theta +\phi )=1 \)
or \( rR=1\)
or \( R=\frac{1}{r}\)
Thus,
\( z^{-1} =\frac{1}{r}( \cos \theta -i\sin \theta )\)

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