finding a complex numberi have no ideea to find a complex number if i know that squared is 3i

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

A complex number has to be found that has a square of 3i.

Let the number be (a + ib)

(a + ib)^2 = 3i

=> a^2 + 2iab + i^2*b^2

=> a^2 - b^2 + 2ab*i

equate the real and imaginary coefficients

2ab = 3

=> a = 3/2*b

substitute in a^2 - b^2 = 0

=> 9/4b^2 = b^2

=> 9 = 4b^4

=> 4b^4 - 9 = 0

=> (2b^2 - 3)(2b^2 + 3) = 0

2b^2 = 3

=> b = sqrt(3/2) and b = -sqrt(3/2)

The other root gives complex values for b and can be ignored.

a = 3/2*b = 3/2*sqrt (3/2) = sqrt (3/2) and a = -sqrt (3/2)

The required complex number is sqrt(3/2) + i*sqrt (3/2) and sqrt -(3/2) - i*sqrt(3/2)

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll write the rectangular form of the complex number z=a+bi.

From enunciation we find that z^2=3i.

We'll raise to square the rectangular form:

z^2=(a+bi)^2

a^2+2abi+(bi)^2=3i

But, from the theory of complex numbers, i^2=-1

a^2+2abi-b^2=3i

Comparing both sides, we'll get:

a^2+b^2=0, a^2=-b^2, so a=b

Also, the imaginary part from the left side has to be equal with the im. part from  the right side.

2ab=3

ab=3/2

But, as we have demonstrated before, a=b

a^2=3/2

a=+/-sqrt6/2

So, the complex number z, whose square is 3i, is:

 z= +/-sqrt6/2(1+i)

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