Question #191f5

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Answer 1 · 2017-04-06T20:21:08

$-4 + 5i to sqrt41(cos(128.7^@) + isin(128.7^@))$

To convert $a + bi to r(cos(theta)+isin(theta))$

$r = sqrt(a^2 + b^2)$

$r = sqrt(-4^2+5^2)$

$r = sqrt(16+25)$

$r = sqrt(41)$

Because "a" is negative and "b" is positive we compensate for the second quadrant by adding $180^@$ :

$theta = tan^-1(b/a)+180^@$

$theta = tan^-1(5/-4)+180^@$

$theta ~~ 128.7^@$

$-4 + 5i to sqrt41(cos(128.7^@) + isin(128.87^@))$