Find the product of the complex numbers z_{1}=\cos\frac{\pi}{6}+i\sin\frac{\pi}{6} and z_{2}=\cos\frac{\pi}{4}+i\sin\frac{\pi}{4} Leave

feminizmiki

feminizmiki

Answered question

2021-12-20

Find the product of the complex numbers z1=cosπ6+isinπ6 and z2=cosπ4+isinπ4
Leave answer in polar form.

Answer & Explanation

Philip Williams

Philip Williams

Beginner2021-12-21Added 39 answers

Step 1
The given two complex numbers are:
z1=cos(π6)+isin(π6)
z2=cos(π4+isin(π4)
we have to find the product of the given two complex numbers, we have to write the answer in polar form.
Step 2
Therefore,
z1z2=(cos(π6)+isin(π6))(cos(π4)+isin(π4))
=cos(π6)cos(π4)+icos(π6)sin(π4)
+isin(π6)cos(π4)+i2sin(π6)sin(π4)
=cos(π6)cos(π4)+i2sin(π6)sin(π4)
+i(cos(π6)sin(π4)+sin(π6)cos(π4))
=cos(π6)cos(π4)+(1)sin(π6)sin(π4)
+i(cos(π6)sin(π4)+sin(π6)cos(π4)) (because i2=1)
Donald Cheek

Donald Cheek

Beginner2021-12-22Added 41 answers

Step 1
The equation complex numbers:
z1×z2=(cos(π6)+isin(π6))×(cos(π4)+isin(π4)):
=2(1+3)4+i2(1+3)4
Answer: 2(1+3)4+i2(1+3)4
RizerMix

RizerMix

Expert2021-12-29Added 656 answers

Step 1
Your input: simplify and calculate different forms of
(cos(π4)+isin(π4))(cos(π6)+isin(π6))
Rewrite:
=(22+2i2)(32+i2)
Use FOIL to multiply (for steps, see foil calculator), don't forget that i2=1
((22+2i2)(32+i2))=(24+64+i(24+64))
Simplify:
(24+64+i(24+64))=(24+64+i(2+6)4)
Hence: =24+64+i(2+6)4
Step 2
Polar form
For a complex number a+bi, polar form given by r(cos(θ)+isin(θ)), where r=a2+b2 and θ=atan(ba)
We have that a=24+64 and =24+64
Thus, r=(24+64)2+(24+64)2=1
Also, θ=atan(24+6424+64)=5π12
Therefore: 24+64+i(24+64)=cos(5π12)+isin(5π12)

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