Answer on Question #45757 – Engineering – Other
Obtain the geometric, polar and exponential representations of ( i 5 − 1 ) − 1 \left(\mathrm{i}5 - 1\right)^{-1} ( i 5 − 1 ) − 1 .
Solution:
z = 1 5 i − 1 = 5 i + 1 ( 5 i − 1 ) ( 5 i + 1 ) = 5 i + 1 − 25 − 1 = 5 i + 1 − 26 = − 1 26 − 5 26 i z = \frac{1}{5i - 1} = \frac{5i + 1}{(5i - 1)(5i + 1)} = \frac{5i + 1}{-25 - 1} = \frac{5i + 1}{-26} = -\frac{1}{26} - \frac{5}{26}i z = 5 i − 1 1 = ( 5 i − 1 ) ( 5 i + 1 ) 5 i + 1 = − 25 − 1 5 i + 1 = − 26 5 i + 1 = − 26 1 − 26 5 i Geometric representation = x + yi
Polar representation:
z = r ( cos φ + i sin φ ) , r = ∣ z ∣ r = x 2 + y 2 = ( − 1 26 ) 2 + ( − 5 26 ) 2 = 1 676 + 25 676 = 26 676 = 1 26 z = 1 26 ( − 1 26 + i ( 25 26 ) ) φ = atan ( − 5 26 − 1 26 ) − π = atan 5 − π \begin{aligned}
& z = r(\cos \varphi + i \sin \varphi), r = |z| \\
r = \sqrt{x^2 + y^2} = \sqrt{\left(-\frac{1}{26}\right)^2 + \left(-\frac{5}{26}\right)^2} = \sqrt{\frac{1}{676} + \frac{25}{676}} = \sqrt{\frac{26}{676}} = \sqrt{\frac{1}{26}} \\
& z = \sqrt{\frac{1}{26}} \left(-\sqrt{\frac{1}{26}} + i \left(\sqrt{\frac{25}{26}}\right)\right) \\
\varphi = \operatorname{atan} \left(\frac{-\frac{5}{26}}{-\frac{1}{26}}\right) - \pi = \operatorname{atan} 5 - \pi \\
\end{aligned} r = x 2 + y 2 = ( − 26 1 ) 2 + ( − 26 5 ) 2 = 676 1 + 676 25 = 676 26 = 26 1 φ = atan ( − 26 1 − 26 5 ) − π = atan 5 − π z = r ( cos φ + i sin φ ) , r = ∣ z ∣ z = 26 1 ( − 26 1 + i ( 26 25 ) ) z = 1 26 ( cos ( atan 5 − π ) + i sin ( atan 5 − π ) ) z = \sqrt {\frac {1}{26}} (\cos (\operatorname {atan} 5 - \pi) + i \sin (\operatorname {atan} 5 - \pi)) z = 26 1 ( cos ( atan 5 − π ) + i sin ( atan 5 − π ))
Exponential representation:
z = z = r e i φ 1 26 e i ( a t a n 5 − π ) z = \frac {z = \mathrm {re} ^ {\mathrm {i} \varphi}}{\sqrt {\frac {1}{26}} \mathrm {e} ^ {\mathrm {i} (\mathrm {atan} 5 - \pi)}} z = 26 1 e i ( atan 5 − π ) z = re i φ
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