Write the complex number z in two algebraic form. z = sqrt2*(cos pi/2 + i sin pi/2)

Expert Answers
hala718 eNotes educator| Certified Educator

z= sqrt2*(cospi/2 + i(sinpi/2)

Let z= a+bi

==> a+bi = sqrt2(c0spi/2) + i(sinpi/2)

But sin pi/2 = 1

cos pi/2 = 0

==> a+ bi = sqrt2( 0 + i*1)

==> a+ bi = 0 + sqrt2*i

==> a= 0

==> b= sqrt2

==> z= sqrt2*i

giorgiana1976 | Student

The given complex number z is given under the trigonometric form:

z = |z|*(cos t+ i*sin t)

where:

|z| = sqrt (a^2 + b^2)

cos t = a / |z| and sint  = b / |z|

We'll identify a,b,|z|:

|z| = sqrt2

cos pi/2 = a/|z|

0 = a/sqrt2

a = 0*sqrt2

a = 0

sin pi/2 = b / |z|

1 = b / sqrt2

b  =sqrt2

Now, we'll put z under the algebraic form:

z = a+b*i, where a is the real part and b is the imaginary part.

z = 0 + i*sqrt2

z = i*sqrt2

neela | Student

The given complex number  Z = sqrt2*(cospi/2 +i sinpi/2)

We know that cospi/2 = 0 and sin pi/2 = 1.

So the given complex number z = sqrt2 {cospi/2 +isinpi/2) =  (sqrt2)(0+i) = (sqrt2)i

Also sqrt2(cospi+isinp/2) = e^(i*pi/2) is another form of writing in the exponent form which is due to Euler.