Write the complex number in polar form.
5 + 2j
5 + 2j
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For any complex number in the form a + bi: convert to polar form r(cos(x) + i sin(x)) as follows.
r = sqrt(a^2 + b^2) and x = arctan(b/a)(from this, you can see that cos(x) = a/r and sin(x) = b/r
arctan(2/5) =~ 21.8 degrees (approx .38 radians)
5 + 2i = sqrt(29)(cos(21.8) + i sin(21.8)), or sqrt(29)cis(21.8)
r = sqrt(a^2 + b^2) and x = arctan(b/a)(from this, you can see that cos(x) = a/r and sin(x) = b/r
arctan(2/5) =~ 21.8 degrees (approx .38 radians)
5 + 2i = sqrt(29)(cos(21.8) + i sin(21.8)), or sqrt(29)cis(21.8)
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¿?¿?¿?¿?¿?, esa fue MI idea!!!
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well well
Polar form: r(cos + i sen) ... then
5+2j --> r = sqrt(25+4)= sqrt (29)
.........Z = sqrt (29)
.........theta = arctan 2/5 = 21.801
Finally Polar form: sqrt (29)[cos 21.801 + i sin 21.801]
simply: sqrt (29)µ21.801
Polar form: r(cos + i sen) ... then
5+2j --> r = sqrt(25+4)= sqrt (29)
.........Z = sqrt (29)
.........theta = arctan 2/5 = 21.801
Finally Polar form: sqrt (29)[cos 21.801 + i sin 21.801]
simply: sqrt (29)µ21.801