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4.49 Complex solutions

Note that the roots in case (iii) are α±i⁢β\alpha\pm i\beta, where by Euler’s formula

e(α+i⁢β)⁢x=eα⁢x⁢ei⁢β⁢x=eα⁢x⁢(cos⁡β⁢x+i⁢sin⁡β⁢x),e^{(\alpha+i\beta)x}=e^{\alpha x}e^{i\beta x}=e^{\alpha x}(\cos\beta x+i\sin% \beta x),
e(α-i⁢β)⁢x=eα⁢x⁢ei⁢β⁢x=eα⁢x⁢(cos⁡β⁢x-i⁢sin⁡β⁢x);e^{(\alpha-i\beta)x}=e^{\alpha x}e^{i\beta x}=e^{\alpha x}(\cos\beta x-i\sin% \beta x);

hence we can write the real solutions from case (iii) as

eα⁢x⁢cos⁡β⁢x=2-1⁢(e(α+i⁢β)⁢x+e(α-i⁢β)⁢x),e^{\alpha x}\cos\beta x=2^{-1}\bigl(e^{(\alpha+i\beta)x}+e^{(\alpha-i\beta)x}% \bigr),
eα⁢x⁢sin⁡β⁢x=(2⁢i)-1⁢(e(α+i⁢β)⁢x-e(α-i⁢β)⁢x).e^{\alpha x}\sin\beta x=(2i)^{-1}\bigl(e^{(\alpha+i\beta)x}-e^{(\alpha-i\beta)% x}\bigr).

Note that ei⁢β⁢xe^{i\beta x} goes round the unit circle as xx increases, while eα⁢xe^{\alpha x} is a real exponential functions such that eα⁢x→∞e^{\alpha x}\rightarrow\infty as x→∞x\rightarrow\infty for α>0\alpha>0, eα⁢x→0e^{\alpha x}\rightarrow 0 as x→∞x\rightarrow\infty for α<0\alpha<0.