Algebra · real student question

Simplify a e^(-4i) + b e^(4i) into trigonometric form using Euler's formula.

Question

Use Euler's formula to write

ae4i+be4ia\,e^{-4i}+b\,e^{4i}

in the form X+iYX+iY with XX and YY real (for real a,ba,b).

Step-by-step solution

  1. State Euler's formula and note that 44 is in radians.

    eiheta=cosθ+isinθe^{i heta}=\cos\theta+i\sin\theta

    Here θ=±4\theta=\pm4 radians - a fixed number, not a variable, so cos4\cos 4 and sin4\sin 4 are just constants (cos4=0.6536\cos4=-0.6536, sin4=0.7568\sin4=-0.7568).

  2. Expand each exponential. Using cos(θ)=cosθ\cos(-\theta)=\cos\theta and sin(θ)=sinθ\sin(-\theta)=-\sin\theta:

    e4i=cos4isin4,e4i=cos4+isin4e^{-4i}=\cos 4-i\sin 4,\qquad e^{4i}=\cos 4+i\sin 4

    The two share the same cosine but opposite sines - that asymmetry is what produces the bab-a below.

  3. Substitute and distribute the coefficients.

    ae4i+be4i=a(cos4isin4)+b(cos4+isin4)a\,e^{-4i}+b\,e^{4i}=a(\cos 4-i\sin 4)+b(\cos 4+i\sin 4)

  4. Group real and imaginary parts. Collecting cos4\cos 4 terms and sin4\sin 4 terms separately:

    =(a+b)cos4+i(ba)sin4=(a+b)\cos 4+i\,(b-a)\sin 4

    So the real part carries the sum a+ba+b and the imaginary part carries the difference bab-a - note the order, bb minus aa.

  5. Check the two special cases. If b=ab=a the imaginary part vanishes and the result is 2acos42a\cos4, matching a(e4i+e4i)=2acos4a(e^{4i}+e^{-4i})=2a\cos4 ✓. If b=ab=-a the real part vanishes, giving 2aisin4-2ai\sin4, matching a(e4ie4i)=2aisin4a(e^{-4i}-e^{4i})=-2ai\sin4 ✓.

  6. Verify numerically. With a=1.3a=1.3, b=0.7b=-0.7: direct evaluation gives 0.392186+1.513605i-0.392186+1.513605i, and (a+b)cos4+i(ba)sin4=0.6(0.65364)+i(2)(0.75680)=0.392186+1.513605i(a+b)\cos4+i(b-a)\sin4=0.6(-0.65364)+i(-2)(-0.75680)=-0.392186+1.513605i ✓.

Answer

ae4i+be4i=(a+b)cos4+i(ba)sin4a\,e^{-4i}+b\,e^{4i}=(a+b)\cos 4+i\,(b-a)\sin 4

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