Since 5i=5e^{i \frac{\pi}{2}} And 2+i=\sqrt5e^{ix_1} where x_1=\arctan \frac12, we have

Russell Gillen 2022-01-03 Answered
Since 5i=5eiπ2 And 2+i=5eix1 where x1=arctan12, we have 5i2+i=5[cos(π2x1)+isin(π2x1)] Now from here how do I continue?
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Answers (4)

Vasquez
Answered 2022-01-08 Author has 460 answers

We already have
LHS=5ei(π2x1),x1=arctan12 (1)
Continue with
RHS=1+2i=5eix2, x2=arctan2 (2)
x1,x2 are complementary angles, x2=π2x1LHS=RHS
Note: polar form represented by exponential form for simplicity in (1)(2). There is no difference between two forms since r and θ are same for LHS and RHS.

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user_27qwe
Answered 2022-01-08 Author has 230 answers

Hint
cos(π2x)=sinx
sin(π2x)=cosx
tan(π2x)=1tanx
So now let's develop the hint from where you left it i.e
5i1+2i=5(cos(π2x)+isin(π2x))=a+ib
So a2+b2=5 and
ba=tan(π2x)=1tanx=2
And so 5a2=5 and a=±1, b=±2 The negative solutions are to be excluded because 0<π2x<π2

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nick1337
Answered 2022-01-08 Author has 575 answers

Using polar form (with exponential notation for convenience),
5i2+i=5etπ25etarctan12=5ei(π2arctan12)=5cos(π5arctan12)+i5sin(π2arctan12)
Sketch the 1,2,5 right triangle and determine that the cosine of the relevant angle (which is complementary to the angle with tangent 12) is 15 and its sine is 25. The result above immediately simplifies to 1+2i.

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nick1337
Answered 2022-01-08 Author has 575 answers

cos(5x)cos(3x)sin(3x)sin(x)=cos(2x)12(cos(2x)+cos(8x))12(cos(2x)cos(4x))=cos(2x)12(cos(4x)+cos(8x))cos(2x)=0cos(8x)+cos(4x)2cos(2x)=02cos(2x)cos(6x)2cos(2x)=0cos(2x)[cos(6x)1]=0cos(2x)=0x=±π4+kπ
discarded because this solution will make zero the denominator of the original equation
cos(6x)=1x=kπ3

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