(-i-5)/(i-6)−i−5i−6
Let me rearrange this
(-i-5)/(i-6)=(-5-i)/(-6+i)=(-(5+i))/(-6+i)=(5+i)/(6-i)−i−5i−6=−5−i−6+i=−(5+i)−6+i=5+i6−i
First of all we have to convert these two numbers into trigonometric forms.
If (a+ib)(a+ib) is a complex number, uu is its magnitude and alphaα is its angle then (a+ib)(a+ib) in trigonometric form is written as u(cosalpha+isinalpha)u(cosα+isinα).
Magnitude of a complex number (a+ib)(a+ib) is given bysqrt(a^2+b^2)√a2+b2 and its angle is given by tan^-1(b/a)tan−1(ba)
Let rr be the magnitude of (5+i)(5+i) and thetaθ be its angle.
Magnitude of (5+i)=sqrt(5^2+1^2)=sqrt(25+1)=sqrt26=r(5+i)=√52+12=√25+1=√26=r
Angle of (5+i)=Tan^-1(1/5)=theta(5+i)=tan−1(15)=θ
implies (5+i)=r(Costheta+isintheta)⇒(5+i)=r(cosθ+isinθ)
Let ss be the magnitude of (6-i)(6−i) and phiϕ be its angle.
Magnitude of (6-i)=sqrt(6^2+(-1)^2)=sqrt(36+1)=sqrt37=s(6−i)=√62+(−1)2=√36+1=√37=s
Angle of (6-i)=Tan^-1((-1)/6)=phi(6−i)=tan−1(−16)=ϕ
implies (6-i)=s(Cosphi+isinphi)⇒(6−i)=s(cosϕ+isinϕ)
Now,
(5+i)/(6-i)5+i6−i
=(r(Costheta+isintheta))/(s(Cosphi+isinphi))=r(cosθ+isinθ)s(cosϕ+isinϕ)
=r/s*(Costheta+isintheta)/(Cosphi+isinphi)*(Cosphi-isinphi)/(Cosphi-isinphi=rs⋅cosθ+isinθcosϕ+isinϕ⋅cosϕ−isinϕcosϕ−isinϕ
=r/s*(costhetacosphi+isinthetacosphi-icosthetasinphi-i^2sinthetasinphi)/(cos^2phi-i^2sin^2phi)=rs⋅cosθcosϕ+isinθcosϕ−icosθsinϕ−i2sinθsinϕcos2ϕ−i2sin2ϕ
=r/s*((costhetacosphi+sinthetasinphi)+i(sinthetacosphi-costhetasinphi))/(cos^2phi+sin^2phi)=rs⋅(cosθcosϕ+sinθsinϕ)+i(sinθcosϕ−cosθsinϕ)cos2ϕ+sin2ϕ
=r/s*(cos(theta-phi)+isin(theta-phi))/(1)=rs⋅cos(θ−ϕ)+isin(θ−ϕ)1
=r/s(cos(theta-phi)+isin(theta-phi))=rs(cos(θ−ϕ)+isin(θ−ϕ))
Here we have every thing present but if here directly substitute the values the word would be tedious for find theta -phiθ−ϕ so let's first find out theta-phiθ−ϕ.
theta-phi=tan^-1(1/5)-tan^-1((-1)/6)θ−ϕ=tan−1(15)−tan−1(−16)
We know that:
tan^-1(a)-tan^-1(b)=tan^-1((a-b)/(1+ab))tan−1(a)−tan−1(b)=tan−1(a−b1+ab)
implies tan^-1(1/5)-tan^-1((-1)/6)=tan^-1(((1/5)-(-1/6))/(1+(1/5)((-1)/6)))⇒tan−1(15)−tan−1(−16)=tan−1⎛⎜⎝(15)−(−16)1+(15)(−16)⎞⎟⎠
=tan^-1((6+5)/(30-1))=tan^-1(11/29)=tan−1(6+530−1)=tan−1(1129)
implies theta -phi=tan^-1(11/29)⇒θ−ϕ=tan−1(1129)
r/s(cos(theta-phi)+isin(theta-phi))rs(cos(θ−ϕ)+isin(θ−ϕ))
=sqrt26/sqrt37(cos(tan^-1(11/29))+isin(tan^-1(11/29)))=√26√37(cos(tan−1(1129))+isin(tan−1(1129)))
=sqrt(26/37)(cos(tan^-1(11/29))+isin(tan^-1(11/29)))=√2637(cos(tan−1(1129))+isin(tan−1(1129)))
This is your final answer.
You can also do it by another method.
By firstly dividing the complex numbers and then changing it to trigonometric form, which is much easier than this.
First of all let's simplify the given number
(5+i)/(6-i)5+i6−i.
Multiply and divide by the conjugate of the complex number present in the denominator i.e 6+i6+i.
(5+i)/(6-i)=((5+i)(6+i))/((6-i)(6+i))=(30+5i+6i+i^2)/(6^2-i^2)5+i6−i=(5+i)(6+i)(6−i)(6+i)=30+5i+6i+i262−i2
=(30+11i-1)/(36-(-1))=(29+11i)/(36+1)=(29+11i)/37=29/37+(11i)/37
(5+i)/(6-i)=29/37+(11i)/37
Let t be the magnitude of (29/37+(11i)/37) and beta be its angle.
Magnitude of (29/37+(11i)/37)=sqrt((29/37)^2+(11/37)^2)=sqrt(841/1369+121/1369)=sqrt(962/1369)=sqrt(26/37)=t
Angle of (29/37+(11i)/37)=Tan^-1((11/37)/(29/37))=tan^-1(11/29)=beta
implies (29/37+(11i)/37)=t(Cosbeta+isinbeta)
implies (29/37+(11i)/37)=sqrt(26/37)(Cos(tan^-1(11/29))+isin(tan^-1(11/29))).
