Question #170360

.Whichofthefolowingistherectangularform ofofthecomplexnumber7.2e

a)0.509+i7.182 b)0.509-2.281c)2.182+i7.217 d)0.217+i7.7.182


1
Expert's answer
2021-03-11T06:19:29-0500

a) z=0.509+i7.182z=0.509+i7.182

z=(0.509)2+(7.182)27.200|z|=\sqrt{(0.509)^2+(7.182)^2}\approx7.200

θ=tan1(7.1820.509)1.500\theta=\tan^{-1}(\dfrac{7.182}{0.509})\approx1.500


z=7.200ei1.500z=7.200e^{i1.500}

b) z=0.509i2.281z=0.509-i2.281

z=(0.509)2+(2.281)22.337|z|=\sqrt{(0.509)^2+(-2.281)^2}\approx2.337

θ=tan1(2.2810.509)1.351\theta=\tan^{-1}(\dfrac{-2.281}{0.509})\approx-1.351


z=2.337ei1.351 or z=2.337ei4.932z=2.337e^{-i1.351}\ or\ z=2.337e^{i4.932}


c) z=2.182+i7.217z=2.182+i7.217

z=(2.182)2+(7.217)27.540|z|=\sqrt{(2.182)^2+(7.217)^2}\approx7.540

θ=tan1(7.2172.182)1.277\theta=\tan^{-1}(\dfrac{7.217}{2.182})\approx1.277


z=7.540ei1.277z=7.540e^{i1.277}

d) z=0.217+i7.182z=0.217+i7.182

z=(0.217)2+(7.182)27.185|z|=\sqrt{(0.217)^2+(7.182)^2}\approx7.185

θ=tan1(7.1820.217)1.541\theta=\tan^{-1}(\dfrac{7.182}{0.217})\approx1.541


z=7.185ei0.1541z=7.185e^{i0.1541}

Answer

a) z=0.509+i7.182=7.200ei1.500z=0.509+i7.182=7.200e^{i1.500}




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