Perform the indicated operations: 1^(°) (1+itanα)/(1-itanα)…
Perform the indicated operations: $$ 1^{\circ}\,\, \frac{1+i\tan\alpha}{1-i\tan\alpha}; \quad 2^{\circ}\,\, \frac{(1-i)^5-1}{(1+i)^5+1}; \quad 3^{\circ}\,\, \frac{(1+2i)^3-(2-i)^3}{(3+2i)^3-(2+i)^2}. $$
Explanation of Method
For $1^{\circ}$, multiply by the conjugate of the denominator and recognize the double-angle formulas for $\cos(2\alpha)$ and $\sin(2\alpha)$. For $2^{\circ}$, compute the powers either via polar form ($1\pm i=\sqrt2\,e^{\pm i\pi/4}$) or by repeated squaring, then rationalize the resulting quotient. For $3^{\circ}$, expand each cube and square directly using $(a+bi)^2$ and $(a+bi)^3=(a+bi)(a+bi)^2$, then rationalize.
Part $1^{\circ}$ · $\dfrac{1+i\tan\alpha}{1-i\tan\alpha}$
Step 1 · Multiply by the conjugate
$$ \frac{1+i\tan\alpha}{1-i\tan\alpha}\cdot\frac{1+i\tan\alpha}{1+i\tan\alpha} = \frac{(1+i\tan\alpha)^2}{(1-i\tan\alpha)(1+i\tan\alpha)} $$
Step 2 · Expand numerator and denominator
$$ (1+i\tan\alpha)^2 = 1+2i\tan\alpha-\tan^2\alpha = (1-\tan^2\alpha)+2i\tan\alpha $$
$$ (1-i\tan\alpha)(1+i\tan\alpha) = 1+\tan^2\alpha $$
Step 3 · Simplify and recognize double-angle identities
$$ \frac{(1-\tan^2\alpha)+2i\tan\alpha}{1+\tan^2\alpha} $$
Using $\cos(2\alpha)=\dfrac{1-\tan^2\alpha}{1+\tan^2\alpha}$ and $\sin(2\alpha)=\dfrac{2\tan\alpha}{1+\tan^2\alpha}$:
Answer for $1^{\circ}$
$$ \boxed{\frac{1+i\tan\alpha}{1-i\tan\alpha} = \cos(2\alpha)+i\sin(2\alpha) = e^{i2\alpha}} $$
i.e. $\text{Re}=\cos(2\alpha)$, $\text{Im}=\sin(2\alpha)$.
Part $2^{\circ}$ · $\dfrac{(1-i)^5-1}{(1+i)^5+1}$
Step 1 · Polar form of $1\pm i$
$$ |1-i|=|1+i|=\sqrt2, \qquad \arg(1-i)=-\frac{\pi}{4}, \quad \arg(1+i)=\frac{\pi}{4} $$
$$ 1-i=\sqrt2\,e^{-i\pi/4}, \qquad 1+i=\sqrt2\,e^{i\pi/4} $$
Step 2 · Raise to the fifth power
$$ (1-i)^5 = (\sqrt2)^5 e^{-i5\pi/4} = 4\sqrt2\left(\cos\tfrac{3\pi}{4}+i\sin\tfrac{3\pi}{4}\right) = 4\sqrt2\left(-\tfrac{\sqrt2}{2}+i\tfrac{\sqrt2}{2}\right) = -4+4i $$
$$ (1+i)^5 = (\sqrt2)^5 e^{i5\pi/4} = 4\sqrt2\left(\cos\tfrac{5\pi}{4}+i\sin\tfrac{5\pi}{4}\right) = 4\sqrt2\left(-\tfrac{\sqrt2}{2}-i\tfrac{\sqrt2}{2}\right) = -4-4i $$
Step 3 · Verify by repeated squaring
$$ (1-i)^2=-2i,\ (1-i)^4=(-2i)^2=-4,\ (1-i)^5=-4(1-i)=-4+4i $$
$$ (1+i)^2=2i,\ (1+i)^4=(2i)^2=-4,\ (1+i)^5=-4(1+i)=-4-4i $$
Both methods agree.
Step 4 · Numerator and denominator of the quotient
$$ (1-i)^5-1 = -5+4i, \qquad (1+i)^5+1 = -3-4i $$
Step 5 · Divide by multiplying by the conjugate
$$ \frac{-5+4i}{-3-4i}\cdot\frac{-3+4i}{-3+4i} = \frac{(-5+4i)(-3+4i)}{(-3-4i)(-3+4i)} $$
$$ (-5+4i)(-3+4i) = 15-20i-12i+16i^2 = -1-32i, \qquad (-3-4i)(-3+4i) = 9-16i^2 = 25 $$
Step 6 · Simplify
$$ \frac{-1-32i}{25} = -\frac{1}{25}-\frac{32}{25}i $$
Answer for $2^{\circ}$
$$ \boxed{\frac{(1-i)^5-1}{(1+i)^5+1} = -\frac{1}{25}-\frac{32}{25}i} $$
Part $3^{\circ}$ · $\dfrac{(1+2i)^3-(2-i)^3}{(3+2i)^3-(2+i)^2}$
Step 1 · Compute $(1+2i)^3$
$$ (1+2i)^2 = 1+4i+4i^2 = -3+4i \quad\Longrightarrow\quad (1+2i)^3=(1+2i)(-3+4i) = -3+4i-6i+8i^2 = -11-2i $$
Step 2 · Compute $(2-i)^3$
$$ (2-i)^2 = 4-4i+i^2 = 3-4i \quad\Longrightarrow\quad (2-i)^3=(2-i)(3-4i) = 6-8i-3i+4i^2 = 2-11i $$
Step 3 · Numerator
$$ (1+2i)^3-(2-i)^3 = (-11-2i)-(2-11i) = -13+9i $$
Step 4 · Compute $(3+2i)^3$ and $(2+i)^2$
$$ (3+2i)^2 = 9+12i+4i^2 = 5+12i \quad\Longrightarrow\quad (3+2i)^3=(3+2i)(5+12i) = 15+36i+10i+24i^2 = -9+46i $$
$$ (2+i)^2 = 4+4i+i^2 = 3+4i $$
Step 5 · Denominator
$$ (3+2i)^3-(2+i)^2 = (-9+46i)-(3+4i) = -12+42i $$
Step 6 · Divide by multiplying by the conjugate
$$ \frac{-13+9i}{-12+42i}\cdot\frac{-12-42i}{-12-42i} = \frac{(-13+9i)(-12-42i)}{(-12+42i)(-12-42i)} $$
$$ (-13+9i)(-12-42i) = 156+546i-108i-378i^2 = 534+438i = 6(89+73i) $$
$$ (-12+42i)(-12-42i) = 144-1764i^2 = 1908 = 36\cdot53 $$
Step 7 · Simplify
$$ \frac{6(89+73i)}{36\cdot53} = \frac{89+73i}{318} = \frac{89}{318}+\frac{73}{318}i $$
Since $318=2\cdot3\cdot53$ and $89$, $73$ are prime, this fraction is already in lowest terms.
Answer for $3^{\circ}$
$$ \boxed{\frac{(1+2i)^3-(2-i)^3}{(3+2i)^3-(2+i)^2} = \frac{89}{318}+\frac{73}{318}i} $$
| Part | Result |
|---|---|
| $1^{\circ}$ | $\cos(2\alpha)+i\sin(2\alpha) = e^{i2\alpha}$ |
| $2^{\circ}$ | $-\dfrac{1}{25}-\dfrac{32}{25}i$ |
| $3^{\circ}$ | $\dfrac{89}{318}+\dfrac{73}{318}i$ |
Result
Key techniques: rationalizing by the conjugate, computing powers of $a+bi$ by repeated squaring or polar form, recognizing double-angle identities, and factoring to simplify final fractions.