Answer: Given : z = r( cosθ + i sin θ ) To Find : \dfrac{z}{\overline{z}} +\dfrac{\overline{z}}{z} z z + z z Solution: z = r( cosθ + i sin θ ) \overline{z} z = r( cosθ – i sin θ ) \dfrac{z}{\overline{z}} +\dfrac{\overline{z}}{z} z z + z z = r( cosθ + i sin θ ) /r ( cosθ – i sin θ ) + r( cosθ – i sin θ ) /r( cosθ + i sin θ ) = ( cosθ + i sin θ ) / ( cosθ – i sin θ ) + ( cosθ – i sin θ ) / ( cosθ + i sin θ ) = ( ( cosθ + i sin θ )² + ( cosθ – i sin θ )²) /(cos²θ – i² sin² θ) = ( 2( cos²θ + i² sin² θ )) /(cos²θ – i² sin² θ) i² = – 1 = ( 2( cos²θ – sin² θ )) /(cos²θ + sin² θ) cos²θ + sin² θ = 1 cos²θ – sin² θ = cos2θ = 2 cos2θ \dfrac{z}{\overline{z}} +\dfrac{\overline{z}}{z} z z + z z = 2 cos2θ Step-by-step explanation: i think this will help u MARK ME AS A BRAINIEST Reply
Answer:
Given : z = r( cosθ + i sin θ )
To Find : \dfrac{z}{\overline{z}} +\dfrac{\overline{z}}{z}
z
z
+
z
z
Solution:
z = r( cosθ + i sin θ )
\overline{z}
z
= r( cosθ – i sin θ )
\dfrac{z}{\overline{z}} +\dfrac{\overline{z}}{z}
z
z
+
z
z
= r( cosθ + i sin θ ) /r ( cosθ – i sin θ ) + r( cosθ – i sin θ ) /r( cosθ + i sin θ )
= ( cosθ + i sin θ ) / ( cosθ – i sin θ ) + ( cosθ – i sin θ ) / ( cosθ + i sin θ )
= ( ( cosθ + i sin θ )² + ( cosθ – i sin θ )²) /(cos²θ – i² sin² θ)
= ( 2( cos²θ + i² sin² θ )) /(cos²θ – i² sin² θ)
i² = – 1
= ( 2( cos²θ – sin² θ )) /(cos²θ + sin² θ)
cos²θ + sin² θ = 1
cos²θ – sin² θ = cos2θ
= 2 cos2θ
\dfrac{z}{\overline{z}} +\dfrac{\overline{z}}{z}
z
z
+
z
z
= 2 cos2θ
Step-by-step explanation:
i think this will help u
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